All questions of Permutation and Combination for GMAT Exam

To Find: Number of ways in which 5 letters can be put in 5 envelopes?

The key to this problem is to avoid list ing all the possibilit ies. Instead, think of an arrangement of five donuts and two dividers. The placement of the dividers determines which man is allotted which donuts, as pictured below: 

In this example, the first man receives one donut, the second man receives three donuts, and the third man receives one donut. Remember that it is possible for either one or two of the men to be allotted no donuts at all. This situation would be modeled with the arrangement below: 

Here, the second man receives no donuts. Now all that remains is to calculate the number of ways in which the donuts and dividers can be arranged: There are 7 objects. The number of ways in which 7 objects can be arranged can be computed by taking 7!:

However, the two dividers are identical, and the five donuts are identical. Therefore, we must divide 7! by 2! and by 5!:

The objective of the question is to find the number of ways in which Amy and Frank can have the leftovers from the pizza party.  

The information given is:

 

Let’s determine the tasks that need to be completed to accomplish the objective: 

Since Frank and Amy want to have one pizza slice and one drink each, they need to fulfil the following tasks:  

Now, in order to accomplish the objective, each of the 4 tasks needs to be completed. Therefore, we will put multiplication signs in the Objective Equation.

The Objective Equation will be:

 

 

Number of ways of selecting 1 item from four different items = ⁴C₁ = 4

Thus, there are 4 ways to do Task 1.

 

Number of ways of selecting 1 item from 3 different items = ³C₁ = 3

Thus, there are 3 ways to do Task 2.

 

Number of ways of selecting 1 item from 2 different items = ²C₁ = 2

Thus, there are 2 ways to do Task 3.

 

Number of ways of selecting 1 item from 1 item = 1

Thus, there is only 1 way to do Task 4.

 

 

(Number of ways in which Frank and Amy can have the leftovers)

  = 4*3*2*1 

  = 24 

So, there are 24 different ways in which Frank and Amy can eat the leftovers from the pizza party. .    

Since order matters in this question (12345 is a different number from 42531), we can solve it using either the Filling Spaces method or the Permutation formula.

The objective of the question here is to find the number of 5-digit numbers that can be formed using the digits 1, 2, 3, 4 and 5 without repeating any digit.

We will solve this question using the Permutation formula.

So, in this step, we will define the single task required to achieve the objective as: arranging 5 digits in 5 spaces

So, the objective equation can be written as:

(Number of 5 digit numbers that can be formed with the digits 1, 2, 3, 4 and 5)

= (Number of ways of arranging 5 digits in 5 spaces) = ⁵P₅

In Step 3, using the Permutation Formula (ⁿP_n = n!), we get that

⁵P₅ = 5!

Now,

5! = 5*4*3*2*1 = 120

So, ⁵P₅ = 120

By putting the value of ⁵P₅ in the objective equation, we get:

(Number of 5 digit numbers that can be formed with the digits 1, 2, 3, 4 and 5) = 120

Looking at the answer choices, we see that Option D is correct.

 The objective of the question is to find the number of ways in which a person can enter the room from one door and exit the room from a different door. 

The information given is:

In order to write the objective equation, we first need to determine the tasks that need to be done in order to accomplish the objective. 

 

Since the person has to enter AND exit the room, the objective comprises of the following tasks:

a. Task 1 – Enter the room

b. Task 2 – Exit the room from a different door 

Now, we need to determine the sign to put between the number of ways of doing each task. Since the objective statement contains the word AND between the two tasks, we will put a multiplication sign in the objective equation.

Thus, our objective equation will be:

 

       a. Task 1 -Enter the room 

Per the information given in the question, there are 6 doors to choose to enter the room.

Thus, there are 6 ways to do Task 1.

      b. Task 2 – Exit the room from a different door 

Once a person has entered the room, he/she has already used one door.

Since different doors have to be used for entry and exit, i.e. repetition is not allowed, only 5 doors are left to exit the door. 

So, there are 5 ways to do Task 2.

 

Now we can plug the values in the objective equation:

    =6 X 5

    = 30

So, there are 30 number of ways in which a person can enter and exit a 6-door room using different doors.

Looking at the answer choices, we see that Option D is the correct answer

To find: Number of ways to distribute 12 different books equally among 4 different boxes.

 

 

 

Looking at the answer choices, we see that the correct answer is Option C

The first thing to recognize here is that this is a permutation with restrictions quest ion. In such questions it is always easiest to tackle the restricted scenario(s) first. The restricted case here is when all of the men actually sit together in three adjacent seats. Restrict ions can often be dealt with by considering the limited individuals as one unit. In this case we have four women ( w1, w2, w3, and w4) and three men (m1, m2, and m3). We can consider the men as one unit, since we can think of the 3 adjacent seats as simply 1 seat. If the men are one unit (m), we are really looking at seating 5 individuals ( w1, w2, w3, w4, and m) in 5 seats. There are 5! ways of arranging 5 individuals in a row. This means that our group of three men is sitting in any of the “five” seats. Now, imagine that the one seat that holds the three men magically splits into three seats. How many different ways can the men arrange themselves in those three seats? 3!. To calculate the total number of ways that the men and women can be arranged in 7 seats such that the men ARE sitting together, we must multiply these two values: 5!3!. However this problem asks for the number of ways the theatre-goers can be seated such that the men are NOT seated three in a row. Logically, this must be equivalent to the following: (Total number of all seat arrangements) – (Number of arrangements with 3 men in a row). The total number of all seat arrangements is simply 7! so the final calculat ion is 7! – 5!3!.

It is important to first note that our point of reference in this quest ion is all the possible subco mmittees that include Michael. We are asked to find what percent of these subcommittees also include Anthony. Let's first find out how many possible subcommittees there are that must include Michael. If Michael must be on each of the three-person committees that we are considering, we are essentially choosing people to fill the two remaining spots of the committee. Therefore, the number of possible committees  can be found by considering the number of different two-people groups that can be formed from a pool of 5 candidates (not 6 since Michael was already chosen). Using the anagram method to solve this combinations question, we assign 5 letters to the various board members in the first row. In the second row, two of the board members get assigned a Y to signify that they were chosen and the remaining 3 get an N, to signify that they were not chosen: 

The number of different combinations of two-person committees from a group of 5 board members would be the number of possible anagrams that could be formed from the word YYNNN = 5! / (3!2!) = 10. Therefore there are 10 possible committees that include Michael. Out of these 10 possible committees, of how many will Anthony also be a member? If we assume that Anthony and Michael must be a member of the three-person committee, there is only one remaining place to fill. Since there are four other board members, there are four possible three-person committees with both Anthony and Michael. Of the 10 committees that include Michael, 4/10 or 40% also include Anthony. The correct answer is C. As an alternate method, imagine splitting the original six-person board into two equal groups of three. Michael is automatically in one of those groups of three. Now, Anthony could occupy any one of the other 5 posit ions -- the 2 on Michael's committee and the 3 on the other committee. Since Anthony has an equal chance of winding up in any of those positions, his chance of landing on Michael's committee is 2 out of 5, or 2/5 = 40%. Since that probabilit y must correspond to the ratio of commit tees asked for in the problem, the answer is achieved. 

Objective:  To find the number of 4-digit passwords.

Information:

 9 digits are available, with no repetition, to form the passwords

 

 This is a classical PERMUTATION problem.

 Objective consists of one task

 Task 1 – Select & Arrange 4 digits from 9 available digits to form passwords.

                              Number of ways to Select & Arrange = Number of 4-digit passwords

 

Since this is a PERMUTATION question, permutation formula applies:

Number of 4-digit passwords = 3024

There are two different approaches to solving this problem. The first employs a purely algebraic approach, as follows: Let us assume there are n teams in a double-eliminat ion tournament. In order to crown a champion, n – 1 teams must be eliminated, each losing exactly two games. Thus, the minimum number of games played in order to eliminate all but one of the teams is 2(n – 1). At the time when the (n – 1)th team is eliminated, the surviving team (the division champion) either has one loss or no losses, adding at most one more game to the total played. Thus, the maximum number of games that can be played in an n-team double-eliminat ion tournament is 2(n – 1) + 1. There were four divisions with 9, 10, 11, and 12 teams each. The maximum number of games that could have been played in order to determine the four division champions was (2(8) + 1) + (2(9) + 1) + (2(10) + 1) + (2(11) + 1) = 17 + 19 + 21 + 23 = 80. The four division champions then played in a single-elimination tournament. Since each team that was eliminated lost exactly one game, the elimination of three teams required exactly three more games. Thus, the maximum number of games that could have been played in order to crown a league champion was 80 + 3 = 83. The correct answer choice is (B). Another way to approach this problem is to use one division as a concrete starting point. Let's think first about the 9-team division. After 9 games, there are 9 losses. Assuming that no team loses twice (thereby maximizing the number of games played), all 9 teams remain in the tournament. After 8 additional games, only 1 team remains and is declared the division winner. Therefore, 9 + 8 = 17 games is the maximum # o f games than can be played in this tournament. We can generalize this information and apply it to the other divisions. To maximize the # of games in the 10-team division, 10 + 9 = 19 games are played. To maximize the # of games in the 11-team division, 11 + 10 = 21 games are played. To maximize the # of games in the 12-team division, 12 + 11 = 23 games are played. Thus, the maximum number of games that could have been played in order to determine the four division champions was 17 + 19 + 21 + 23 = 80. After 3 games in the single elimination tournament, there will be 3 losses, thereby eliminating all but the one championship team. Thus, the maximum number of games that could have been played in order to crown a league champion was 80 + 3 = 83. Once again.

 The objective of the question is to find the number of ways in which Marie can select ONE package of apples for her apple pie. 

 

The information given is:

In order to write the objective equation, we first need to determine the tasks that need to be done to accomplish the objective.

Since Marie has to select ONE apple package for her pie, she can do the following tasks: 

a. Task 1 – Select a package of fresh apples            

b. Task 2 – Select a package of canned apples 

  Now that we know the tasks, we need to determine whether to put a multiplication or an addition sign in the objective equation.                         

In order to accomplish the objective, Marie can select either a fresh apple package or a canned apple package. Since doing either of the two tasks will suffice, we will put an addition sign in the objective equation.

Therefore, the objective equation will be formed as follows:

 

      a. Task 1 – Select a package of fresh apples            

There are two varieties of fresh apples.

This means, there are 2 ways to do Task 1.

b.   Task 2 – Select a package of canned apples

There are three brands of canned apples.

This means, there are 3 ways to do Task 2.

 

In this step, we will plug the values in the objective equation:

   = 3 + 2

     = 5

So, there are 5 different ways in which Marie can select one package of apples for her recipe.                                                                                                                                                    

Looking at the answer choices, we see that Option C is correct.

Approach

The objective of the question is to find the number of ways in which a person can get a prime number on the first dice OR a composite number on second dice.              

There is not much information provided to us in the question itself. However, we can deduce a few things: 

So, let’s move on to the next step in which we’ll write the objective equation.

Achievement of the objective involves two tasks:

Task 1: Getting a prime number on the first dice

Task 2: Getting a composite number on the second dice

Since the objective statement contains the word 'OR' between the two tasks, we will put an addition sign between the two tasks. The objective equation will therefore be:

(Number of ways to achieve the objective) = (Number of ways to get a prime number on the first dice) + (Number of ways to get a composite number on the second dice)

Now that we have the objective equation, let’s move on to the next step to find the number of ways in which these tasks can be completed.  

There are three prime numbers in the numbers from 1 to 6. 

Thus, there are 3 ways to do Task 1.

There are total 2 composite numbers in the range 1-6.  

So, there are 2 ways to do Task 4. 

(Number of ways to achieve the objective) = 3 + 2 = 5

 

So, there are 5 different ways in which a person can get a prime number on the first dice or a composite number on the second dice.  

 

To find: Number of different triangles that may be specified.

 

 

Looking at the answer choices, we see that the correct answer is Option B

The 10 civil engineers can be arranged in a round table in 
(10-1)! = 9! Ways ---(A)
Now we need to arrange software engineers the round table such that software engineers 
and civil engineers are seated alternatively. i.e., we can arrange 10 software engineers
in the 10 positions marked as * as shown below

This can be done in 10P10 = 10! Ways ---(B)

From (A) and (B),
The required number of ways = 9! × 10!

The easiest way to solve this quest ion is to consider the restrictions separately. Let’s start by considering the restriction that one of the parents must drive, temporarily ignoring the restriction that the two sisters won't sit next to each other. This means that… 
2 people (mother or father) could sit in the driver’s seat 4 people (remaining parent or one of the children) could sit in the front passenger seat 3 people 0could sit in the first back seat 2 people could sit in the second back seat 1 person could sit in the remaining back seat.
The total number of possible seat ing arrangements would be the product of these various possibilit ies: 2 × 4 × 3 × 2 × 1 = 48. We must subtract from these 48 possible seating arrangements the number of seating arrangements in which the daughters are sitting together. The only way for the daughters to sit next to each other is if they are both sitting in the back. This means that…  2 people (mother or father) could sit in the driver’s seat 2 people (remaining parent or son) could sit in the front passenger seat Now for the back three seats we will do something a little different. The back three seats must contain the two daughters and the remaining person (son or parent). To find out the number of arrangements in which the daughters are sitting adjacent, let’s consider the two daughters as one unit. The remaining person (son or parent) is the other unit. Now, instead of three seats to fill, we only have two "seats," or units, to fill.
There are 2 × 1 = 2 ways to seat these two units. However, the daughter-daughter unit could be d1d2 or d2d1 We must consider both of these possibilities so we multiply the 2 by 2! for a total of 4 seating possibilities in the back. We could also have manually counted these possibilities: d1d2X, d2d1X, Xd1d2, Xd2d1 Now we must multiply these 4 back seat scenarios by the front seat scenarios we calculated earlier: 
(2 × 2) × 4 = 16 front back If we subtract these 16 "daughters-sitting-adjacent" scenarios from the total number of "parent-driving" scenarios, we get: 48 – 16 = 32 

There are two possibilit ies in this problem. Eit her Kim and Deborah will both get chocolate chip cookies or Kim and Deborah will both get oatmeal cookies. If Kim and Deborah both get chocolate chip cookies, then there are 3 oatmeal cookies and 2 chocolate chip cookies left for the remaining four children. There are 5!/3!2! = 10 ways for these 5 remaining cookies to be distributed--four of the cookies will go to the children, one to the dog. (There are 5! ways to arrange 5 objects but the three oatmeal cookies are identical so we divide by 3!, and the two chocolate chip cookies are identical so we divide by 2!.) If Kim and Deborah both get oatmeal cookies, there are 4 chocolate chip cookies and 1 oatmeal cookie left for the remaining four children. There are 5!/4! = 5 ways for these 5 remaining cookies to be distributed--four of the cookies will go to the children, one to the dog. (There are 5! ways to arrange 5 objects but the four chocolate chip cookies are identical so we divide by 4!.) Accounting for both possibilities, there are 10 + 5 = 15 ways for the cookies to be distributed.

Ignoring Frankie's requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be either ahead of or behind Joey. Logically, since the combinat ions favor neither Frankie nor Joey, each would be behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie's requirement, the six mobsters could be arranged in 720/2 = 360 different ways.

To Find: Number of possible codes?

4.  Case-II: If the letter selected is a consonant

5.  Total number of ways of forming a code = Number of codes formed in Case-I + Number of codes formed in Case-II

Number of ways in which 1 letter and 2 digits can be arranged = 3!

 

 

To Find: Number of ways to select the Director and the President?

Hence, there are 23 ways in which the company can select its director and the president.

 

The objective of the question is to find the number of ways in which three different letters can be SELECTED from a set of 8 different letters: A, B, S, O, L, U, T and E   

Note that only selection is to be considered here. We don’t need to arrange these letters since the question did not ask us to make words.  

This is a Selection question. Order doesn’t matter here. So we will solve it by using the ⁿC_r formula.

In this case, only one task needs to be completed to accomplish the objective: Select 3 letters out of the given set of 8 different letters.

So, the objective equation becomes:

(Number of ways of selecting 3 letters out of 8 different letters) = ⁸C₃

By using the formula:

We get,

 Upon simplifying this expression we get 

                        ⁸C₃=56

 

 Step 4: Calculate the final answer

 By putting this value in the objective equation, we get:

(Number of ways of selecting 3 letters out of 8 different letters) = 56

Here n = 5, k = 7. Hence, as per the above formula, the required number of ways

This is a relat ively simple problem that can be fiendishly difficult unless you have a good approach to solving it and a solid understanding of how to count. We will present two different strategies here. Strategy 1: This problem seems difficult, because you need to figure out how many distinct orientations the cube has relative to its other sides. Given that you can rotate the cube in an unlimited number of ways, it is very difficult to keep track of what is going on – unless you have a system. Big hint: In order to analyze how mult iple things behave or compare or are arranged relative to each other, the first thing one should do is pick a reference point and fix it. Here is a simple example. Let’s say you have a round table with four seat positions and you want to know how many distinct ways you can orient 4 people around it relative to each other (i.e., any two orientations where all 4 people have the same person to their left and to their right are considered equivalent). Let’s pick person A as our reference po int and anchor her to the North position. Think about this next statement and convince yourself that it is true: By choosing A as a fixed reference, all distinct arrangements of the other 3 people relative to A will constitute the complete set of distinct arrangements of all 4 people relative to each other. Hence, fixing the location of one person makes it significantly easier to keep track of what is going on. Given A is fixed at the North, the 3 other people can be arranged in the 3 remaining seats in 3! = 6 ways, so there are 6 dist inct orientations of 4 people sitt ing around a circular table. Using the same principle, we can conclude that, in general, if there are N people in a circular arrangement, after fixing one person at a reference point, we have (N-1)! distinct arrangements relative to each other. Now let’s solve the problem. Assume the six sides are: Top (or T), Bottom (or B), N, S, E, and W, and the six colors are designated 1, 2, 3, 4, 5, and 6. Following the first strategy, let’s pick color #1 and fix it on the Top side of the cube. If #1 is at the Top position, then one of the other 5 colors must be at the Bottom position and each of those colors would represent a distinct set of arrangements. Hence, since there are exactly 5 possible choices for the color of the Bottom side, the number of unique arrangements relative to #1 in the Top position is a multiple of 5. For each of the 5 colors paired with #1, we need to arrange the other 4 colors in the N, S, E, and W positions in distinct arrangements. Well, this is exactly like arranging 4 people around a circular table, and we have already determined that there are (n-1)! or 3! ways to do that. Hence, the number of distinct patterns of painting the cube is simply 5 x 3! = 30. 
Strategy 2: There is another way to solve this kind of problem. Given one dist inct arrangement or pattern, you can try to determine how many equivalent ways there are to represent that one particular arrangement or pattern within the set of total permutations, then divide the total number of permutations by that number to get the number of distinct arrangements. This is best illustrated by example so let’s go back to the 4 people arranged around a circular table. Assume A is in the North posit ion, then going clockwise we get B, then C, then D. Rotate the table 1/4 turn clockwise. Now we have a different arrangement where D is at the North position, followed clockwise by A, then B, then C. BUT, this is merely a rotation of the distinct relative position of the 4 people (i.e., everyone still has the same person to his right and to his left) so they are actually the same arrangement. We can quickly conclude that there are 4 equivalent or non-dist inct arrangements for every dist inct relat ive posit ioning of the 4 people. We can arrange 4 people in a total of 4! = 24 ways. However, each DISTINCT arrangement has 4 equivalents, so in order to find the number of distinct arrangements, we need to divide 4! by 4, which yields 3! or 6 distinct ways to arrange 4 people around a circular table, the same result we got using the “fixed reference” method in Strategy 1. Generalizing, if there are N! ways to arrange N people around a table, each dist inct relat ive rotation can be represented in N ways (each 1/Nth rotation around the table) so the number of dist inct arrangements is N!/N = (N-1)! Now let’s use Strategy #2. Consider a cube that is already painted in a particular way. Imagine putting the cube on the table, with color #1 on the top side. Note, that by rotating the cube, we have 4 different orientations of this particular cube given color #1 is on top. Using symmetry, we can repeat this analysis when #1 is facing any of the other 5 directions. Hence, for each of the six directions that the side painted with #1 can face, there are 4 ways to orient the cube. Consequently, there are 6 x 4 = 24 total orientations of any one cube painted in a particular manner. Since there are 6 sides and 6 colors, there are 6! or 720 ways to color the six sides each with one color. However, we have just calculated that each DISTINCT pattern has 24 equivalent orientations, so 720 must be divided by 24 to get the number of distinct patterns. This yields 720/24 = 30, confirming the answer found using Strategy #1.

The objective of the question is to find the number of ways in which three different letters can be SELECTED from a set of 4 letters: T, H, R and E.

(Please note that the word THREE has 5 alphabets, but of these, E occurs twice. So, we get only 4 different alphabets from the word THREE: T, H, R and E)

This is a Selection question. Order doesn’t matter here. So we will solve it by using the ⁿC_r formula.

In this case, only one task needs to be completed to accomplish the objective: Select 3 letters out of the given set of 4 different letters.

So, the objective equation becomes:

(Number of ways of selecting 3 letters out of 4 different letters) = ⁴C₃

By using the formula:

We get:

  Upon simplifying this expression, you get 

                                                     ⁴C₃=4

 Step 4: Calculate the final answer

 By putting this value in the objective equation, we get:

(Number of ways of selecting 3 letters out of 4 different letters) = 4

 

To find: Number of ways to constitute a 4 – member panel, such that all members have different professions

(RequiredNumberofways)=

 

 

 

 

 

 

So, since we have not been able to determine a unique number of professions, Statement 1 alone is not sufficient.

 

We quickly see that multiple values for the number of professions and for the number of people in each profession are possible here.

So, Statement 2 is not sufficient to answer the question

 

Since we now know the number of professions and the number of candidates in each profession, we will be able to answer the Q.

The 2 statements together are sufficient to answer the question.

 

Answer: Option C

 

The objective of the question is to find the number of ways in which a room can be illuminated.  

The information given is:

                             

Since it is told that the room is illuminated even if at least one bulb is switched on, we need to fulfil the following tasks to accomplish the objective:  

                                

Now, in order to accomplish the objective a person can complete any of the above mentioned tasks. And hence he needs perform only of these actions. So, principle of addition will be used to form objective equation as follows:

In this step, we’ll find out how many ways are there to perform various steps:

Number of ways of selecting one items from 7 different items = ⁷C₁ = 7

Thus, there are 7 ways to do Task 1.

 

Number of ways of selecting two items from 7 different items = ⁷C₂ = 21 

Thus, there are 21 ways to do Task 2.  

 

Number of ways of selecting three items from 7 different items = ⁷C₃ = 35 

Thus, there are 35 ways to do Task 3. 

 

Number of ways of selecting 4 items from 7 different items = ⁷C₄ = 35 

Thus, there are 35 ways to do Task 4. 

 

Number of ways of selecting 5 items from 7 different items = ⁷C₅ = 21 

Thus, there are 21 ways to do Task 5. 

 

Number of ways of selecting 6 items from 7 different items = ⁷C₆ = 7  

Thus, there are 7 ways to do Task 6. 

 

Number of ways of selecting 7 items from 7 different items = ⁷C₇ = 1 

Thus, there is only one way to do Task 7. 

 

              

(Number of ways to illuminate the room) = 7 + 21 + 35 + 35 + 21 + 7 +1  

                                                                              = 127 

So, there are 127 different ways in which the room can be illuminated.      

The objective of the question is to list down unique two-digit integers or two-letter words.  

 The information that can be deduced from the given question is:  

 

Since Leslie has to make a two-digit numbers or a two-letter words, the objective comprises of the following:  

Now, in order to accomplish the objective, Leslie can write use either 2-digits OR 2-alphabets.

Therefore, we will put an addition sign in the Objective Equation as under:

       

 

Now we can plug the values in the objective equation:

Given: 4-letter word GALE

To find: Number of 3-letter words that can be formed without repeating any letters

We will use the method of filling spaces to answer the question

 

Answer: Option E

First, let's consider the different medal combinations that can be awarded to the 3 winners: (1) If there are NO TIES then the three medals awarded are: GOLD, SILVER, BRONZE. (2) What if there is a 2-WAY tie? --If there is a 2-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, SILVER. --If there is a 2-WAY tie for SECOND, then the medals awarded are: GOLD, SILVER, SILVER. --There cannot be a 2-WAY tie for THIRD (because exactly three medals are awarded in total). (3) What if there is a 3-WAY tie? --If there is a 3-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, GOLD. --There are no other possible 3-WAY ties. Thus, there are 4 possible medal combinations: (1) G, S, B (2) G, G, S (3) G, S, S (4) G, G, G. Now let's determine how many different ways each combination can be distributed. We'll do this by considering four runners: Albert, Bob, Cami, and Dora.
COMBINATION 1: Gold, Silver, Bronze

Therefore, there are different victory circles that will contain 1 GOLD, 1 SILVER, and 1 BRONZE medalist. COMBINATION 2: Gold, Gold, Silver.
Using the same reasoning as for Combination 1, we see that there are 24 different victory circles that will contain 2 GOLD medalists and 1 SILVER medalist. However, it is important to realize that these 24 victory circles must be reduced due to "overcounting." To illustrate this, consider one of the 24 possible Gold-GoldSilver victory circles: Albert is awarded a GOLD. Bob is awarded a GOLD. Cami is awarded a SILVER. Notice that this is the exact same victory circle as the following: Bob is awarded a GOLD. Albert is awarded a GOLD. Cami is awarded a SILVER. Each victory circle has been "overcounted" because we have been counting each different ordering of the two gold medals as a unique victory circle, when, in reality, the two different orderings consist of the exact same victory circle. Thus, the 24 victory circles must be cut in half; there are actually only 12 unique victory circles that will contain 2 GOLD medalists and 1 SILVER medalist. (Note that we did not have to worry about "overcounting" in Combinat ion 1, because each of those 24 possibilit ies was unique.) 
COMBINATION 3: Gold, Silver, Silver.
Using the same reasoning as for Combination 2, we see that there are 24 possible victory circles, but only 12 unique victory circles that contain 1 GOLD medalist and 2 SILVER medalists.
COMBINATION 4: Gold, Gold, Gold.
Here, once again, there are 24 possible victory circles. However, because all three winners are gold medalists, there has been a lot of "overcounting!" How much overcounting? Let's consider one of the 24 possible GoldGold-Gold victory circles: Albert is awarded a GOLD. Bob is awarded a GOLD. Cami is awarded a GOLD. 
Notice that this victory circle is exact ly the same as the following victory circles: Albert-GOLD, Cami-GOLD, Bob-GOLD. Bob-GOLD, Albert-GOLD, Cami-GOLD. Bob-GOLD, Cami-GOLD, Albert-GOLD. CamiGOLD, Albert-GOLD, Bob-GOLD. Cami-GOLD, Bob-GOLD, Albert-GOLD. Each unique victory circle has actually been counted 6 times! Thus we must divide 24 by 6 to find the number of unique victory circles. There are actually only unique victory circles that contain 3 GOLD medalists. FINALLY, then, we have the following: 
(Combination 1) 24 unique GOLD-SILVER-BRONZE victory circles. (Combination 2) 12 unique GOLDGOLD-SILVER victory circles. (Combination 3) 12 unique GOLD-SILVER-SILVER victory circles. (Combination 4) 4 unique GOLD-GOLD-GOLD victory circles. Thus, there are unique victory circles.

 

 

 

 

There is a total of four numbers: 0, 2, 4 and 6

Similar to the previous problem, to accomplish the objective of forming 3-digit numbers, three task needs to be done:

Since we have a restriction of not using ‘0’ in the hundreds digit, we should first deal with that restriction.

Therefore,

·       Task1: Selecting a number for hundreds digit

o   This task can be achieved in 3 ways.

(We can choose from any 3 numbers: 2, 4 or 6. As we need a 3-digit number we cannot use ‘0’ at this place)

·       Task2: Selecting a number for tens digit

o   This task can be achieved in 3 ways.

(We can choose among the remaining 3 numbers since we already selected one for the hundreds digit)

·       Task3: Selecting a number for units’ digit

o   This task can be achieved in 2 ways.

(We can choose among the remaining 2 numbers since we already selected two: one for the hundreds digit and other for the tens digit)

 

·       All these three tasks must be done to form three-digit number.

 

Therefore,

·       The objective of forming unique 3-digit numbers can be found by multiplying the number of ways in which each task is achieved.

 

Therefore,

Correct Answer: Option C

 

We can also use the permutation formula (nP_r) to find our answer.

Out of the given four numbers (0, 2,4,6) we have to arrange three numbers so as to make a unique 3-digit number.

We also know that we cannot use 0 as hundreds digit.

For the time being, let us consider that ‘0’ can be the hundreds digit. 

If we also include ‘0’ as hundreds digit, then we are simply arranging three numbers from a list of four numbers.

Now we know that among these 24 unique numbers, some numbers are those with ‘0’ as the hundreds digit. So, we need to subtract those.

If ‘0’ is fixed as the hundreds digit, then for the tens and units digit (2 places) we have the option to choose from remaining 3 numbers (2,4,6).

 

Therefore,

Therefore,

Correct Answer : Choice C

To find: Number of ways to choose the 2 teams for a Tennis Mixed Doubles Match

So, Number of ways in which the players can be selected

= (Ways to Select 2 men out of 5)*(Ways to select 2 women out of 4)*(Ways to arrange the 2 selected men and 2 selected women into 2 teams)

 

 

Looking at the answer choices, we see that the correct answer is Option D

To find: Number of positive 4-digit numbers that are divisible by 20

 

 

Looking at the answer choices, we see that the correct answer is Option C

The objective of the question is to find the number of routes to go from Tempe (Friend’s home) to Peoria (Work) via Dunlap (Tom’s home). That is, the objective of the question is to go from Tempe to Dunlap AND to go from Dunlap to Peoria.

The information given is:

 

Based on the above information, we can draw the following diagram: 

 

In order to write the objective equation, we first need to determine the tasks that need to be done to accomplish the objective.

The objective here comprises of following tasks:

     a.   Task 1 – Go from Tempe (Friend’s Home) to Dunlap (Tom’s Home)

 b.   Task 2 – Go from Dunlap (Tom’s Home) to Peoria (Work)

                                    

Now, let’s look at the objective statement again:

Since the objective equation has an AND between the two tasks, we will put a multiplication sign between the number of ways of doing each task.

Therefore, the objective equation will be:

a. Task 1 -Go from Tempe (Friend’s Home) to Dunlap (Tom’s Home)

Per the information given in the question, there are 8 routes between Tempe and Dunlap.

Thus, there are 8 ways to do Task 1.

 

   2.   Task 2 -Go from Dunlap (Tom’s Home) to Peoria (Work)

Per the information given in the question, there are 5 routes from Dunlap to Peoria.

 Thus, there are 5 ways to do Task 2.

 

In this step, we plug the values obtained in Step 3 in the objective equation:

= 8  X 5

= 40

So, there are 40 different routes by which Tom can go from Tempe (Friend’s Home) to Peoria (Work).

Looking at the answer choices, we see that Option E is the correct answer.

As we have learnt the first step of solving any permutation and combination question is to understand the objective of the question. So, let’s do that!

The objective of the question is to find out the total number of different outcomes possible if 6 different coins are tossed at the same time.  

The adjective ‘different’ used before the word ‘coins’ is of crucial importance here. Since all coins are different, this means that getting Heads on the 1^st coin and Tails on all the others is a different outcome from getting Heads on the 2^nd coin and Tails on all the others. (Note: there are many ways in which the 6 coins may be different. They may each be of a different denomination, or they may each of different size or color etc.)

In order to write the objective equation, we first need to determine all the tasks that need to be done to accomplish the objective. 

Since the person has to toss 6 coins at the same time, the objective comprises of six tasks, as listed below:

Now, in order to accomplish the objective, all the six tosses need to be performed. So, we will put a multiplication sign between the number of ways of doing each task.

Therefore, the objective equation will be written as under:

 

Now we can plug the values in the objective equation. We get:

2*2*2*2*2*2 = 64

 So, 64 different outcomes are possible if 6 different coins are tossed simultaneously.  

Since order matters in this question (12340 is a different number from 42310), we can solve it using either the Filling Spaces method or the Permutation formula.

The objective of the question here is to find the number of 5-digit numbers that can be formed using the digits 0, 1, 2, 3 and 4 without repeating any digit.

One thing to note in this question is that one of the digits is 0. When the 5 digits are arranged such that the first digit is 0 (for example: 01234), the result is a 4-digit number. The question specifically asks about the number of five digit numbers only. This means that after considering all the possible permutations of the given 5 digits, we need to subtract the cases in which the first digit is 0 (which makes the number effectively a 4-digit number)

Now, in the numbers where the first digit is 0, there are 4 digits left (1, 2, 3 and 4) to be arranged in 4 spaces.

So,

 

This is the objective equation.

Now, we know that

The number of ways in which 5 digits can be arranged in 5 spaces = ⁵P₅

And

The number of ways in which 4 digits can be arranged in 4 spaces = ⁴P₄

So, the objective equation becomes:

(Number of 5-digit numbers that can be formed using the digits 0, 1, 2, 3 and 4)

= ⁵P₅ -⁴P₄

In Step 3, using the Permutation Formula (ⁿP_n = n!), we get that

⁵P₅ = 5! = 5*4*3*2*1 = 120

And,

⁴P₄ = 4! = 4*3*2*1 = 24

 

By putting these valuesin the objective equation, we get:

(Number of 5-digit numbers that can be formed using the digits 0, 1, 2, 3 and 4)

= 120 – 24 = 96

Looking at the answer choices, we see that Option C is correct.

Let W be the number of wins and L be the number of losses. Since the total number of hands equals 12 and the net winnings equal $210, we can construct and solve the following simultaneous equations: w + l = 12, 100 w – 10 l = 210. so l = 9, w = 3. So we know that the gambler won 3 hands and lost 9. We do not know where in the sequence of 12 hands the 3 wins appear. So when counting the possible outcomes for the first 5 hands, we must consider these possible scenarios: 1) Three wins and two losses 2) Two wins and three losses 3) One win and four losses 4) No wins and five losses In the first scenario, we have WWWLL. We need to know in how many different ways we can arrange these five letters: 5!/2!3! = 10. So there are 10 possible arrangements of 3 wins and 2 losses. The second scenario --WWLLL -- will yield the same result: 10. The third scenario -- WLLLL -- will yield 5 possible arrangements, since the one win has only 5 possible positions in the sequence. The fourth scenario --LLLLL -- will yield only 1 possible arrangement, since rearranging these letters always yields the same sequence. Altogether, then, there are 10 + 10 + 5 + 1 = 26 possible outcomes for the gambler's first five hands.

This is a counting problem that is best solved using logic. First, let’s represent the line of women as follows:
0000
0000
where the heights go from 1 to 8 in increasing order and the unknowns are designated 0s. Since the women are arranged by their heights in increasing order from left to right and front to back, we know that at a minimum, the lineup must conform to this:
0008
1000
Let’s further designate the arrangement by labeling the other individuals in the top row as X, Y and Z, and the individuals in the bottom row as A, B, and C.
XYZ8
1ABC
Note that Z must be greater than at least 5 numbers (X, Y, B, A, and 1) and less than at least 1 number (8). This means that Z can only be a 6 or a 7. Note that Y must be greater than at least 3 numbers (X, A and 1) and less than at least 2 numbers (8 and Z). This means that Y can only be 4, 5, or 6. Note that X must be greater than at least 1 number (1) and less than at least 3 numbers (8, Z and Y), This means that X must be 2, 3, 4, or 5. This is enough information to start counting the total number of possibilities for the top row. It will be easiest to use the middle unknown value Y as our starting point. As we determined above, Y can only be 4, 5, or 6. Let’s check each case, making our conclusions logically: If Y is 4, Z has 2 options (6 or 7) and X has 2 options (2 or 3). This yields 2 x 2 = 4 possibilities. If Y is 5, Z has 2 options (6 or 7), and X has 3 options (2, 3, or 4). This yields 2 x 3 = 6 possibilities. If Y is 6, Z has 1 option (7), and X has 4 options (2, 3, 4, or 5). This yields 1 x 4 = 4 possibilities. For each of the possibilities above, the bottom row is completely determined because we have 3 numbers left, all of
which must be in placed increasing order. Hence, there are 4 + 6 + 4 = 14 ways for the women to pose.

 Given: Rectangle ABCD in which AB is parallel to the x-axis and AD is parallel to the y-axis

To find: Number of squares of side 1 unit that can be drawn inside or on the rectangle ABCD

Let’s find the Number of squares of side 1 unit that can be formed in the same row as square APQR (let’s call this number r):

​

Similarly, we can find that the Number of squares that can be formed in the same column as square APQR (let’s call this number c) =

if AD is an integer

 

(Total number of squares that can be formed in the rectangular region ABCD) = r*c

 

Since the values of r and c depend on AB and AD, we need to find the values of AB and AD, that is, the length and breadth of rectangle ABCD.

Statement 1 says that ‘The length of side AB is 6 units’

 

Statement 2 says that ‘The coordinates of points A and C are (3,2) and (9, 5)’

 

 

 

Since we’ve already arrived at a unique answer in Step 4, this step is not required

Answer: Option B

heparin is used for preventing  blood clot

Plasminogen is the inactive precursor to plasmin so, it's out of scenario  it's inactive not gonna do anything 

plasmin main hero it  dissolves the blood clot , it's not preventing not an inactive precursor it's working as blood clot dissolver 

so here is our answer.

PDGF It's help in making blood clot when we cut our finger from knife it's PDGF( Platelet derived growth factor) which help to stop our blood . it's totally out of scenario.

so, plasmin is the one who is really working as blood clot dissolver.

LCM(x, y) =

To Find: Number of values y can take?

, but it can take only one value of prime factor 3, which is 3⁴

4. So, number of possible values of y = Number of ways in which 2 can be present in y * Number of ways in which 3 can be present in y * Number of ways in which 5 can be present in y* Number of ways in which 7 can be present in y

 and 

2. Prime factors for which their powers in LCM (x, y) is greater than that in x

 and 7 do not have the same powers in LCM (x, y) and x

3. Prime factors for which their powers in LCM (x, y) and x are same

a. 3 and 5 have the same powers in both LCM (x, y) and x

, i.e

a total of (2b+c+1) ways………(3)

2. Possible number of ways in which 5 can be present in y = 

, i.e. a total of (c+1) ways…….(4)

4. So, number of possible values of y = 1 * 1 * (2b+c+1) * (c+1) = (2b+c+1) (c+1)

So, there can be (2b+c+1) (c+1) values of y.

Since order matters in this question (Seats B, C and D occupied with Jill, Greta and Pauline is a different seating arrangement from Seats B, C and D occupied with Greta, Pauline and Jill), we can solve it using either the Filling Spaces method or the Permutation formula.

There are 7 people in this question – 1 boy and 6 girls.

And, there are 7 chairs.

Each chair is unique, because it is marked with a different number.

In this question, two cases are possible:

Case 1: The boy sits on Chair A and the 6 girls are arranged on chairs B-G

Case 2: The boy sits on Chair G and the 6 girls are arranged on chairs A-F

The question here wants us to find the total number of seating arrangements in which the boy sits either on the chair A or on chair G. That is, the question wants us to find the total number of ways in which the 6 girls can be arranged either on chairs B-G or on chairs A-F. This is the objective of the question.

The objective here consists of two tasks:

Next, we need to determine what sign to put in the objective equation – multiplication or addition.

Let’s look at the objective statement again:

The objective statement contains the words ‘Either. . . Or’

This means that we will put an addition sign between the number of ways of doing the two tasks.

Thus, the objective equation will be:

 

Now, we know that

The number of ways in which 6 girls can be arranged in 6 seats = ⁶P₆

So, the objective equation becomes:

                            (Number of arrangements in which the boy sits on chirs A or G) = 6P₆ + 6P₆

That is,

                            (Number of arrangements in which the boy sits on chirs A or G) = 2 (6P₆)

In Step 3, using the Permutation Formula (ⁿP_n = n!), we get that

⁶P₆ = 6! = 6*5*4*3*2*1 = 720

                     

By putting these values in the objective equation, we get:

                              (Number of arrangements in which the boy sits on chirs A or G) = 2 x720 = 1440

 

 Looking at the answer choices, we see that Option C is correct.