Test: Permutations and Combinations- 2 - CA Foundation MCQ

Step 1: Total number of letters in "MONDAY"
The word "MONDAY" has 6 distinct letters: M, O, N, D, A, Y.

Step 2: Fixing the first letter as 'M'
Since the word must begin with 'M', we can fix 'M' in the first position. Now we are left with 5 positions to arrange the remaining letters: O, N, D, A, Y.

Step 3: Calculate the number of arrangements without any restriction
If there were no restriction on the last letter, we would arrange the remaining 5 letters (O, N, D, A, Y) in the remaining 5 positions. The number of ways to do this is:
5! = 5 × 4 × 3 × 2 × 1 = 120

Step 4: Subtract the cases where the word ends with 'N'
Now, we need to subtract the cases where the word ends with 'N'. If 'N' is at the last position, we are left with arranging the remaining 4 letters: O, D, A, Y.
The number of ways to arrange these 4 letters is:
4! = 4 × 3 × 2 × 1 = 24

Step 5: Calculate the total number of valid arrangements
The total number of arrangements where the word starts with 'M' and does not end with 'N' is:
5! - 4! = 120 - 24 = 96

Final Answer:
The number of arrangements in which the letters of the word "MONDAY" can be arranged such that the word begins with 'M' and does not end with 'N' is 96.

We are asked to find the number of ways in which the letters of the word "MOBILE" can be arranged such that consonants always occupy the odd places.

Step 1: Identify the Consonants and Vowels
The word "MOBILE" consists of 6 letters: M, O, B, I, L, E.
Consonants: M, B, L
Vowels: O, I, E
So, there are 3 consonants (M, B, L) and 3 vowels (O, I, E).

Step 2: Odd Positions in the Word
The odd positions in the 6-letter word are 1, 3, and 5.

Step 3: Arrangement of Consonants in Odd Positions
We need to arrange 3 consonants (M, B, L) in 3 odd positions (1, 3, 5). The number of ways to do this is:
3! = 3 × 2 × 1 = 6

Step 4: Arrangement of Vowels in the Remaining Positions
After placing the consonants in the odd positions, the 3 vowels (O, I, E) will occupy the remaining positions (2, 4, 6). The number of ways to arrange these vowels is:
3! = 3 × 2 × 1 = 6

Step 5: Total Number of Arrangements
The total number of ways to arrange the letters is the product of the two independent arrangements (consonants in odd positions and vowels in even positions):
3! × 3! = 6 × 6 = 36
    
Final Answer:
The number of ways in which the letters of the word "MOBILE" can be arranged such that consonants always occupy the odd places is 36.

We are asked to find the number of ways in which 5 persons can sit at a round table such that the tallest person is always to the right of the shortest person.

Step 1: Total Number of Arrangements without Any Restrictions
In a round table arrangement, the number of ways to arrange n persons is given by (n - 1)! because the arrangement is circular (rotations of the same arrangement are considered identical).

For 5 persons, the total number of ways to arrange them without any restriction is:
(5 - 1)! = 4! = 4 × 3 × 2 × 1 = 24

Step 2: Fixing the Positions of the Tallest and Shortest Persons
Since the tallest person must always be on the right side of the shortest person, we need to ensure that this condition is satisfied.
In a circular arrangement, if we fix the position of the shortest person (since the table is round, we can consider one position as fixed), there is only one specific seat to the right of the shortest person where the tallest person can sit. So, the tallest person's seat is fixed once we fix the shortest person's position.

Step 3: Arranging the Remaining 3 Persons
Once the shortest and tallest persons are seated, the remaining 3 persons can be arranged in the remaining 3 positions. The number of ways to arrange these 3 persons is:
3! = 3 × 2 × 1 = 6

Step 4: Final Calculation
Since we have fixed the positions of the shortest and tallest persons and can arrange the other 3 persons in 6 ways, the total number of valid arrangements is: 6
Final Answer:
The number of ways to arrange 5 persons at a round table such that the tallest person is always on the right side of the shortest person is 6.

We are asked to find the value of ¹²C₄ + ¹²C₃.

Step 1: Formula for Combinations
The formula for combinations is:
ⁿC_k = n! / (k!(n - k)!)

Step 2: Calculate ¹²C₄
Using the formula for combinations, we calculate ¹²C₄:
¹²C₄ = 12! / (4!8!) = (12 × 11 × 10 × 9) / (4 × 3 × 2 × 1) = 495

Step 3: Calculate ¹²C₃
Similarly, we calculate ¹²C₃:
¹²C₃ = 12! / (3!9!) = (12 × 11 × 10) / (3 × 2 × 1) = 220

Step 4: Add ¹²C₄ and ¹²C₃
Now, we add ¹²C₄ and ¹²C₃:
¹²C4 + ¹²C₃ = 495 + 220 = 715

Final Answer:
The value of 12C4 + 12C3 is 715.

We are given the equation ¹⁸C_r = ¹⁸C_(r+2) and are asked to find the value of ^rC₅.

Step 1: Understand the given equation
The given equation is:
¹⁸Cr = ¹⁸C_(r+2)

Using the property of combinations, we know that nCk = nC(n-k). Therefore, we can equate the two combinations:
¹⁸C_r = ¹⁸C_(18-r-2) (since r + 2 = 18 - r)
This simplifies to:
r = 18 - r - 2

Step 2: Solve for r

Now, solving for r:
r + r = 18 - 2
2r = 16
r = 8

Step 3: Find the value of ^rC₅
Now that we know r = 8, we need to find ⁸C₅.
We use the combination formula:
⁸C₅ = 8! / (5!3!) = (8 × 7 × 6) / (3 × 2 × 1)

Now, simplifying the factorials:
⁸C₅ = 336 / 6 = 56

Final Answer:
The value of ^rC₅ is 56.

We are given the following equations:

We need to find the value of r.

Step 1: Use the property of binomial coefficients
The property of binomial coefficients states:

Substituting the values of 

This simplifies to:

Step 2: Solve for n
Now, multiply both sides by 2r:
r=2(n−(r−1))
Simplifying:
r=2n−2r+2
Now, move all terms involving r to one side:
3r=2n+2
Now, we have an equation relating r and n.
Step 3: Use another property of binomial coefficients
The property of binomial coefficients also states:

Substituting the values of 

This simplifies to:

Step 4: Solve for n
Now, cross-multiply:
2(r+1)=7(n−r)
Simplifying:
2r+2=7n−7r
Move all terms involving r to one side:
9r=7n−2

Step 5: Solve the system of equations
We now have two equations:

• 3r=2n+2
• 9r=7n−2

Solving this system of equations, we get 
r=6.
Therefore, the value of r is 6.

For each friend, the person has two choices: either to invite them or not to invite them. Since there are 8 friends, each with 2 choices, the total number of ways to choose friends (including choosing none) is:

2⁸ = 256
However, since the question specifies "one or more" friends, we subtract the case where no friends are invited.
256-1 = 255

Correct answer: b) 255

We are given four electrical appliances: T.V, Refrigerator, Washing Machine, and Cooler.

The problem asks us to find the number of ways a person can choose one or more appliances from these four.

Step 1: Understanding the choices

For each appliance, a person can either:

• Choose it, or
• Not choose it.

So, for each of the 4 appliances, there are 2 choices: choose or not choose.

Step 2: Total number of choices

For all 4 appliances, the total number of choices is:

2 * 2 * 2 * 2 = 16

But, this count includes the case where no appliance is chosen (i.e., choosing none of the appliances). Since we need at least one appliance to be chosen, we subtract the case where no appliances are chosen.

Step 3: Subtract the case of choosing none

The number of choices where no appliance is chosen is 1 (i.e., the empty set). So, we subtract this from the total number of choices:

16 - 1 = 15

The number of ways in which a person can choose one or more appliances is 15.

We are given the equation: ⁿC₁₀ = ⁿC₁₄

We need to find the value of ²⁵C_n.

Step 1: Use the property of binomial coefficients

We know the property of binomial coefficients:

ⁿC_r = ⁿC_(n-r)

From the given equation nC10 = nC14, we can apply the property:

ⁿC₁₀ = ⁿC_(n-10)

So, we have:

10 = n - 14

Solving for n:

n = 24

Step 2: Find the value of ²⁵C_n

Now, we need to find ²⁵C_n where n = 24:

²⁵C₂₄ = 25

The value of ²⁵C_n is 25.

We are given 7 gents and 4 ladies, and a committee of 5 members is to be formed. The condition is that each committee must include at least one lady. We need to find the number of such committees.

Step 1: Total number of committees
First, calculate the total number of committees that can be formed with no restrictions. The total number of ways to select 5 people from 11 (7 gents and 4 ladies) is given by the combination formula:
Total committees = 11C5 = (11 * 10 * 9 * 8 * 7) / (5 * 4 * 3 * 2 * 1) = 462

Step 2: Committees with no ladies
Next, calculate the number of committees that can be formed with no ladies (i.e., all gents). This is simply selecting 5 gents from 7:
Committees with no ladies = 7C5 = (7 * 6) / (2 * 1) = 21

Step 3: Committees with at least one lady
The number of committees with at least one lady is found by subtracting the number of committees with no ladies from the total number of committees:
Committees with at least one lady = Total committees - Committees with no ladies = 462 - 21 = 441
The number of committees that include at least one lady is 441.

We are given the equation:

28C_2r : 24C_2r – 4 = 225 : 11

Step 1: Express the binomial coefficients
We can express the binomial coefficients 28C_2r and 24C_2r as:
28C_2r = ²⁸C_2r and 24C_2r = ²⁴C_2r
Using the properties of combinations, we can substitute values for the left-hand side of the equation.

Step 2: Set up the equation
The equation simplifies to:

28C_2r / 24C_2r = 225 / 11 + 4

Step 3: Simplify the right-hand side
225 / 11 + 4 = 225 / 11 + 44 / 11 = 269 / 11

Step 4: Solve for r
Now, we can solve the equation and find the value of r:
28C_2r / 24C_2r = 269 / 11

Using the property of combinations and solving for r:
r = 7
The value of r is 7.

Hint: The number of diagonals in a polygon of n sides is 1/2n (n–3).

Explanation

- Given that there are 12 points in a plane.
- Out of these 12 points, 5 are collinear.
- To find the number of triangles that can be formed using these 12 points, we can use the combination formula.
- The number of ways to choose 3 points out of 12 is given by 12C3 = 220.
- However, we need to subtract the number of triangles that can be formed using the collinear points.
- Since 5 points are collinear, we can form triangles using these points as well.
- The number of ways to choose 3 points out of 5 collinear points is given by 5C3 = 10.
- So, the total number of triangles that can be formed using the 12 points is 220 - 10 = 210.
- Therefore, the correct answer is option C: 210.

To find the number of straight lines that can be formed by 16 non-collinear points in a plane, we can use the concept of combinations. Here’s the step-by-step solution:

Step 1: Understanding the Problem
We need to find the number of straight lines that can be formed by joining any two points from a set of 16 non-collinear points. Since no three points are collinear, any two points will form a unique straight line.

Step 2: Using Combinations
To determine how many ways we can choose 2 points from the 16 points, we use the combination formula:

In this case, n=16 (the total number of points) and r=2 (the number of points we are choosing to form a line).

Step 3: Applying the Combination Formula
Substituting the values into the formula gives us:

Step 5: Calculating 2!
Calculating 2!:

2!=2×1=2

Step 6: Final Calculation
Now substituting back into our equation:

The total number of handshakes in a party where each person shakes hands with every other person can be calculated using the formula for combinations, since the handshakes involve selecting 2 people from a total of n people. The number of ways to choose 2 people from n people is given by:

Number of handshakes = ι_n(ι_n-1)/2

We are told that the total number of handshakes is 66. Therefore, we can set up the equation:

n(n - 1)/2 = 66

Multiplying both sides by 2 to eliminate the fraction:

n(n - 1) = 132

Now, solve for n. This is a quadratic equation:

n² - n - 132 = 0

To solve this quadratic equation, we can use the quadratic formula:

n = (-(-1) ± √((-1)² - 4(1)(-132)))/2(1)

n = (1 ± √(1 + 528))/2

n = (1 ± √529)/2

n = (1 ± 23)/2

So, n = (1 + 23)/2 = 12 or (1 - 23)/2 = -11.

Since the number of guests cannot be negative, we have n = 12.

To form a parallelogram, we need to select two lines from one set of parallel lines and two lines from the other set. The intersection of these lines will give the four corners of the parallelogram.

Step 1: Number of ways to choose 2 lines from 4 parallel lines
This can be calculated using the combination formula ι_n^r, where n is the total number of lines and r is the number of lines we want to choose. So, the number of ways to choose 2 lines from 4 is:

ι₄² = (4 × 3) / (2 × 1) = 6

Step 2: Number of ways to choose 2 lines from 3 parallel lines
Similarly, the number of ways to choose 2 lines from 3 is:

ι₃² = (3 × 2) / (2 × 1) = 3

Step 3: Total number of parallelograms
Now, to form a parallelogram, we multiply the number of ways to choose 2 lines from each set:

Total number of parallelograms = ι₄² × ι₃² = 6 × 3 = 18

We are asked to find the number of ways in which 12 students can be equally divided into three groups. This is a problem of distributing the students into three groups where the groups are indistinguishable, and each group has 4 students.

Step 1: Total ways to divide students into 3 groups
The number of ways to divide 12 students into 3 groups of 4 students each is given by the formula for partitioning n objects into r groups of equal size:

Ways = ¹²C₄ × ⁸C₄ × ⁴C₄ / 3!

Step 2: Calculation of combinations
First, we calculate the individual combinations:

¹²C₄ = (12 × 11 × 10 × 9) / (4 × 3 × 2 × 1) = 495

⁸C₄ = (8 × 7 × 6 × 5) / (4 × 3 × 2 × 1) = 70

⁴C₄ = 1

Step 3: Apply the formula
Now, we apply the formula to find the total number of ways:

Ways = (495 × 70 × 1) / 6 = 5775

We are asked to find the number of chords that can be formed by joining 8 points marked on the circumference of a circle. A chord is formed by selecting two distinct points on the circle.

Step 1: Choosing 2 points from 8
To form a chord, we need to select 2 points from the 8 points marked on the circle. The number of ways to choose 2 points from 8 is given by the combination formula:
Number of chords = ι₈²

Step 2: Apply the combination formula
The combination formula for choosing 2 points from 8 is:
ι₈² = (8 × 7) / (2 × 1) = 28

We are tasked with forming a committee of 3 ladies and 4 gents from 8 ladies and 7 gents, with the restriction that Mrs. X refuses to serve in a committee if Mr. Y is also a member.

Step 1: Total number of committees without restrictions
First, let's calculate the total number of committees that can be formed without any restrictions. The committee consists of 3 ladies and 4 gents, so we can choose 3 ladies from 8 and 4 gents from 7. This can be calculated as:

Number of ways to choose 3 ladies from 8 = ι₈³
Number of ways to choose 4 gents from 7 = ι₇⁴

Step 2: Apply the combination formula
Now, let's compute the values using the combination formula:
ι₈³ = (8 × 7 × 6) / (3 × 2 × 1) = 56
ι₇⁴ = (7 × 6 × 5 × 4) / (4 × 3 × 2 × 1) = 35

So, the total number of committees without restrictions is:
56 × 35 = 1960

Step 3: Committees where Mrs. X and Mr. Y are both members
Now, let's consider the cases where both Mrs. X and Mr. Y are on the committee. If Mrs. X is on the committee, we need to choose 2 more ladies from the remaining 7 ladies (since one lady, Mrs. X, is already chosen) and 3 more gents from the remaining 6 gents (since Mr. Y is already chosen). This can be calculated as:

Number of ways to choose 2 ladies from 7 = ι₇²
Number of ways to choose 3 gents from 6 = ι₆³

Step 4: Apply the combination formula
ι₇² = (7 × 6) / (2 × 1) = 21
ι₆³ = (6 × 5 × 4) / (3 × 2 × 1) = 20

So, the number of committees where both Mrs. X and Mr. Y are included is:
21 × 20 = 420

Step 5: Calculate the final number of valid committees
The number of valid committees (where Mrs. X and Mr. Y are not both members) is the total number of committees without restrictions minus the number of committees where both Mrs. X and Mr. Y are on the committee:

Valid committees = 1960 - 420 = 1540

The problem states that the Supreme Court has given a 6 to 3 decision upholding a lower court. We need to find the number of ways the court can give a majority decision reversing the lower court.

Step 1: Understanding the Majority Decision
The total number of justices is 9 (6 upheld and 3 reversed). For a majority decision reversing the lower court, at least 5 justices must favor the reversal. We need to calculate how many ways we can choose 5 justices who will reverse the lower court.

Step 2: Applying the Combination Formula
We have 9 justices in total, and we need to select 5 justices who will reverse the decision. The number of ways to choose 5 justices from 9 is given by the combination formula:
ι₉⁵ =   n! / (r! (n - r)!)

Step 3: Calculation
Now, calculate ι₉⁵:
ι₉⁵ = (9 × 8 × 7 × 6 × 5) / (5 × 4 × 3 × 2 × 1) = 126
Thus, the number of ways to select 5 justices who will reverse the decision is 126.

The correct answer is: Option A: 256

We are given a scenario where there are five bulbs, out of which three are defective. These bulbs are to be tested in two bulb points in a dark room, and we need to find the number of trials required to light the room.

Step 1: Understanding the Setup

• There are 5 bulbs in total.
• 3 of them are defective (meaning they do not light up).
• 2 bulbs will be tested at a time in the bulb points.
• We need to determine how many trials (pairs of bulbs) are required to light the room.

Step 2: Conditions for Lighting the Room
For the room to be lit, at least one of the bulbs in each pair must be non-defective. Therefore, to ensure the room lights up, the two bulbs selected must either have:

• Two non-defective bulbs (both will light up).
• One non-defective bulb and one defective bulb (the non-defective one will light up).

Step 3: Counting Possible Combinations of Trials
Total number of bulbs = 5 (3 defective and 2 non-defective).
The number of ways to select 2 bulbs from the 5 bulbs is given by the combination formula:

ι₅² = π / (2 × 1)

Step 4: Subtracting the Invalid Trials
The total number of valid trials is:
Valid trials = ι₅² - ι₃² = 10 - 3 = 7
The number of trials where the room will be lit is 7.

We are asked to find the ratio of the number of arrangements of the letters of the words "CALCUTTA" and "AMERICA."

Step 1: Number of Arrangements of CALCUTTA
The word "CALCUTTA" consists of 8 letters in total: C, A, L, C, U, T, T, A.
The repeated letters are:

• "C" appears 2 times,
• "A" appears 2 times,
• "T" appears 2 times.

The number of distinct arrangements of the letters of "CALCUTTA" is given by the formula for permutations of a multiset:
Arrangements of CALCUTTA = π / (2! × 2! × 2!)

Calculating:
Arrangements of CALCUTTA = (8!) / (2! × 2! × 2!) = 40320 / 8 = 5040

Step 2: Number of Arrangements of AMERICA
The word "AMERICA" consists of 7 letters: A, M, E, R, I, C, A.
The letter "A" repeats 2 times.
The number of distinct arrangements of the letters of "AMERICA" is:
Arrangements of AMERICA = π / (2!)

Calculating:
Arrangements of AMERICA = (7!) / (2!) = 5040 / 2 = 2520

Step 3: Finding the Ratio
The ratio of the number of arrangements of "CALCUTTA" to "AMERICA" is:
Ratio = (5040) / (2520) = 2:1
Thus, the ratio of the number of arrangements is 2 : 1.

We are asked to find the number of ways of selecting 4 letters from the word "EXAMINATION".

Step 1: Understanding the Letters in EXAMINATION
The word "EXAMINATION" consists of 11 letters: E, X, A, M, I, N, A, T, I, O, N.
The frequency of each letter is as follows:

• E appears 1 time
• X appears 1 time
• A appears 2 times
• M appears 1 time
• I appears 2 times
• N appears 2 times
• T appears 1 time
• O appears 1 time

Step 2: Casework Based on the Number of Repeated Letters
We need to select 4 letters, so we will calculate the cases based on how many repeated letters we choose.

Case 1: No letter repeats (all letters are different)
The number of ways to choose 4 different letters from 8 distinct letters (E, X, A, M, I, N, T, O) is:
Number of ways = ι₈⁴ = π / (4! × 1!)

Case 2: One letter repeats twice, and two other letters are different
We can choose 1 letter that repeats 2 times (A, I, or N) and then select 2 other different letters from the remaining 7 distinct letters. The number of ways for this case is:

Number of ways = 3 (choices for the repeated letter) × ι₇² (ways to select 2 other letters)

Case 3: Two letters repeat twice each
We can choose 2 letters that each repeat 2 times (A, I, or N). The number of ways for this case is:
Number of ways = ι₃² = 3

Step 3: Total Number of Ways
Adding all the cases together, we get the total number of ways to select 4 letters:
Total ways = 70 + 56 + 6 = 136
The total number of ways to select 4 letters from the word "EXAMINATION" is 136.

• Selecting and Arranging the Consonants:

  • Selection: Choose 4 consonants out of 12.
  • Arrangement: Arrange the selected 4 consonants in order.
  • Number of Ways: This is represented by the permutation notation ¹²P₄.

• Selecting and Arranging the Vowels:

  • Selection: Choose 3 vowels out of 5.
  • Arrangement: Arrange the selected 3 vowels in order.
  • Number of Ways: This is represented by the permutation notation ⁵P₃.

• Combining Consonants and Vowels:

  • After selecting and arranging the consonants and vowels separately, we need to arrange all 7 letters together to form a word.
  • Number of Ways: The 7 letters can be arranged in 7! (7 factorial) ways.

• Total Number of Different Words:

  • Multiply the number of ways to arrange consonants, vowels, and the combined letters.

  • Total Ways = ¹²P₄ × ⁵P₃ × 7! = 4950 x 7!

We are asked to find the number of ways in which 8 guests can be seated, with 4 on each side of a long rectangular table. Additionally, 2 particular guests desire to sit on one side of the table, and 3 on the other side.

Step 1: Understanding the Problem
There are 8 guests in total. - 2 particular guests want to sit on one side, and 3 other particular guests want to sit on the other side. - Each side of the table can accommodate 4 guests.

Step 2: Arrangements on the First Side (with 2 Particular Guests)
On the first side, we must first place the 2 particular guests in 2 of the 4 seats. - The number of ways to select 2 seats for the particular guests is given by the combination formula:

Ways to choose 2 seats = ι₄² = 4! / (2!2!) = 6
After selecting the 2 seats for the particular guests, we can arrange them in these 2 seats in 2! ways.

Ways to arrange the 2 particular guests = 2! = 2
Now, there are 2 remaining seats for the other 2 guests. The remaining 2 guests can be seated in these 2 seats in 2! ways.

Ways to arrange the other 2 guests = 2! = 2
Thus, the total number of ways to arrange guests on the first side is:

Total for first side = 6 × 2! × 2! = 6 × 2 × 2 = 24

Step 3: Arrangements on the Second Side (with 3 Particular Guests)
On the second side, we have 3 particular guests who must occupy 3 of the 4 seats. - The number of ways to choose 3 seats for the particular guests is given by the combination formula:

Ways to choose 3 seats = ι₄³ = 4! / (3!1!) = 4
The 3 particular guests can be arranged in these 3 seats in 3! ways.

Ways to arrange the 3 particular guests = 3! = 6
The remaining 1 seat can be occupied by the last remaining guest in 1 way. Thus, the total number of ways to arrange guests on the second side is:

Total for second side = 4 × 6 × 1 = 24

Step 4: Total Number of Arrangements
The total number of seating arrangements is the product of the arrangements for both sides:

Total arrangements = 24 (for the first side) × 24 (for the second side) = 576
The total number of seating arrangements is 1728.