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All questions of Combinations and Permutations for ACT Exam

How many three digit odd numbers can be formed by using the digits 5,6,7,8 if the repetition of digits is allowed?
  • a)
    64
  • b)
    32
  • c)
    24
  • d)
    None of these
Correct answer is 'B'. Can you explain this answer?

Pooja Shah answered
The unit place can be filled in 2 ways because question asked to find 3 digit odd numbers from 5,6,7,8 and the other two places can be filled in 4 ways as repetition is allowed.
So, Unit's place can be filled in 2 ways. Decimal place can be filled in 4 ways. Hundred's place can be filled in 4 ways.
 => 4*4*2 = 32.

A lady wants to select one cotton saree and one polyster saree from a textile shop. If there are 15 cotton and 13 polyster varieties in that shop, in how many ways can she pick up two sarees?
  • a)
    125
  • b)
    215
  • c)
    345
  • d)
    195
Correct answer is option 'D'. Can you explain this answer?

Suresh Reddy answered
The lady can select one cotton saree out of 15 cotton varieties in 15 ways since
any of 15 varieties can be selected. Corresponding to each selection of a cotton saree, she can
choose a polyester saree in 13 ways. Hence the two sarees (one cotton and one polyester), by
multiplication principle of counting, can be selected in 15 x 13= 195 ways

Nidhi has 6 friends. In how many ways can she invite one or more of them to a party at her home?
  • a)
    64
  • b)
    63
  • c)
    25
  • d)
    24
Correct answer is option 'B'. Can you explain this answer?

Lavanya Menon answered
She has 6 friends and he wants to invite one or more. That is the same as saying he wants to invite at least 1 of his friends.
 
So, the number of ways he could do this is:
Invite only one friend
Invite any two friends
Invite any three friends
Invite any four friends
Invite any five friends
Invite all six friends
This can be thought of in terms of combinations. Inviting  r  friends out of  n  is same as choosing  r  friends out of  n . So, we can write the possibilities as:
6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 
= 6 + 15 + 20 + 15 + 6 + 1 
= 63

There are 7 chairs in a row. In how many ways can 3 persons occupy any three of them
  • a)
    240
  • b)
    210
  • c)
    260
  • d)
    120
Correct answer is option 'B'. Can you explain this answer?

Neha Joshi answered
First person can sit on any of the 7 seats, second person can sit on other 6 vacant seats except one occupied by the first one. Similarly, the third can do sit on 5 so, total no. of ways = 7*6*5 = 210

A lady arranges a dinner party for 6 guests .The number of ways in which they may be selected from among 10 friends if 2 of the friends will not attend the party together is
  • a)
    164
  • b)
    140
  • c)
    112
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?

Neha Joshi answered
Let us say that the two particular friends are A and B.
If A is invited among six guests and B is not, then:  number of  combinations to select 5 more guests from the remaining 8 friends:
          C(8, 5) =  8 ! / (5! 3!)  = 56
If B is invited among the six guests and A is not , then the number of ways of selecting the remaining 5 guests =  C(8, 5) =  56
Suppose both A and B are not included in the six guests list : then the number of such combinations =  C(8, 6) = 7 * 8 /2 = 28
So the total number of sets of guests that can be selected =  140.

Number of signals that can be made using 2 flags out of given 4 flags.
  • a)
    3
  • b)
    6
  • c)
    12
  • d)
    4
Correct answer is option 'C'. Can you explain this answer?

Om Desai answered
No. of ways of selecting two flags out of four = 4C2
So, total possible different signals generated =  4C2×2!
⟹ 6×2=12

 In how many ways can 3 letters be posted in 4 letter boxes?
  • a)
    27
  • b)
    4!
  • c)
    64
  • d)
    3!
Correct answer is option 'C'. Can you explain this answer?

Preeti Iyer answered
The first letter can be posted in 4 ways. So, total outcomes about the first letter = 4.
For every outcome about the first letter, the second letter can be posted in 4 ways. So, total outcomes about the first and the second letters= (4*4) = 16.
Therefore, following the same route, we can say, total possible outcomes about the first and the second letters = (4*4*4) = 64.

How many three digit odd numbers can be formed by using the digits 1,2,3,4,5,6 if the repetition of digit is not allowed:
  • a)
    216
  • b)
    120
  • c)
    60
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Lavanya Menon answered
There can be only 3 digits at the last digit (1,3,5) the middle digit should have 5 No's. out of 6, because one digit is used in last digit similarly 1st digit will have 4 No's.
Therefore : 4×5×3 => 60

In how many ways can a cricket team be selected out of 15 players if a particular player has to be included?
  • a)
    C(14,11)
  • b)
    C(14,10)
  • c)
    C(15,11)
  • d)
    C(15,10)
Correct answer is option 'A'. Can you explain this answer?

Pooja Shah answered
11 players can be selected out of 16 in 16C11 ways = 16!/(11! 5!) = 4368 ways. Now, If two particular players is to be included and one particular player is to be rejected, then we have to select 9 more players out of 13 in 13C9 ways.
= 13!/(9! 4!) = 715 ways.

The number of different ways in which a man can invite one or more of his 6 friends to dinner is?
  • a)
    30
  • b)
    63
  • c)
    120
  • d)
    15
Correct answer is option 'B'. Can you explain this answer?

Naina Bansal answered
He can invite one or more friends by inviting 1 friend, or 2 friends or 3 friends, or all the 6 friends.

1 friend can be selected out of 6 in 6C1 = 6 ways

2 friends can be selected out of 6 in 6C2 = 15 ways

3 friends can be selected out of 6 in 6C3 = 20 ways

4 friends can be selected out of 6 in 6C4 = 15 ways

5 friends can be selected out of 6 in 6C5 = 6 ways

6 friends can be selected out of 6 in 6C6 = 1 ways

Therefore the required number of ways (combinations) = 6 + 15 + 20 + 15 + 6 + 1 = 63

How many numbers are there between 100 and 1000 , in which all the digits are distinct?
  • a)
    729
  • b)
    648
  • c)
    858
  • d)
    546
Correct answer is option 'B'. Can you explain this answer?

Nandini Patel answered
H T U

9.9.8 = 9 x 9 x 8 = 648 numbers

No. of digits in hundreds place should be filled first as the zero should be excluded

(1,2,3,4,5,6,7,8,9) = 9 ways

No. of digits in tens place (now the zero can be included but one digit has been used at hundreds place) so, no. of ways we can fill ten's place is = 9 ways

No of digits in units place is any one among the remaining 8 no. = 8 ways

so, 9 x 9 x 8 = 648

A coin is tossed 6 times, in how many throws can 4 heads and 2 tails be obtained?
  • a)
    24
  • b)
    18
  • c)
    15
  • d)
    10
Correct answer is option 'C'. Can you explain this answer?

Naina Bansal answered
Whether we toss a coin 6 times or six coins one time the number of arrangement will remain same .

As to find number of ways we get 4 heads and 2 tails out of 6 times  

The number of different ways in which a man can invite one or more of his 6 friends to dinner is
  • a)
    63
  • b)
    15
  • c)
    30
  • d)
    120
Correct answer is option 'A'. Can you explain this answer?

Om Desai answered
He can invite one or more friends by inviting 1 friend, or 2 friends or 3 friends, or all the 6 friends.
1 friend can be selected out of 6 in 6C1 = 6 ways
2 friends can be selected out of 6 in 6C2 = 15 ways
3 friends can be selected out of 6 in 6C3 = 20 ways
4 friends can be selected out of 6 in 6C4 = 15 ways
5 friends can be selected out of 6 in 6C5 = 6 ways
6 friends can be selected out of 6 in 6C6 = 1 ways
Therefore the required number of ways (combinations) = 6 + 15 + 20 + 15 + 6 + 1 = 63

The number of ways in which 6 “ + “ and 4 “ – “ signs can be arranged in a line such that no two “ – “ signs occur together is
  • a)
    P(10,4)
  • b)
    C(7,4)
  • c)
    C(10,4)
  • d)
    none of these
Correct answer is option 'C'. Can you explain this answer?

Poonam Reddy answered
′+′ signs can be put in a row in one way creating seven gaps shown as arrows:
Now 4′−′ signs must be kept in these gaps. So, no tow ′−′ signs should be together.
Out of these 7 gaps 4 can be chosen in 7C4 ways.

How many 4 digit numbers are there, when a digit may be repeated any number of times?
  • a)
    3×43
  • b)
    9×103
  • c)
    104
  • d)
    410
Correct answer is option 'B'. Can you explain this answer?

Nandini Patel answered
If we assume the leading digit can not be zero, then the first 4-digit number is 1000.
The last 4-digit number is 9999.
The number of 4-digit numbers is therefore:
9999 - 1000 + 1 = 9000

——
Another way to calculate this: 
1..9 are available for the first digit —> 9 choices
0..9 are available for the 2nd, 3rd, and 4th digits —> 10 choices for each: 10*10*10 = 1000 choices total
and 9*1000 = 9000

If nC8 = nC2, then n is
  • a)
    8
  • b)
    12
  • c)
    10
  • d)
    2
Correct answer is option 'C'. Can you explain this answer?

Explanation:

To solve this problem, we need to use the formula for the combination (nCr), which is given by:

nCr = n! / (r!(n-r)!)

where n is the total number of objects, and r is the number of objects chosen at a time.

Given that nC8 = nC2, we can set up the equation as follows:

n! / (8!(n-8)!) = n! / (2!(n-2)!)

To simplify the equation, we can cancel out the common terms on both sides:

(n-2)! * 8! = (n-8)! * 2!

Now, let's expand the factorials:

(n-2)! * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = (n-8)! * 2 * 1

Cancel out the common terms:

8 * 7 * 6 * 5 * 4 * 3 = (n-8)!

We are left with:

8 * 7 * 6 * 5 * 4 * 3 = (n-8)!

Now, let's evaluate the factorial on the left side:

8 * 7 * 6 * 5 * 4 * 3 = 20160

So, we have:

20160 = (n-8)!

To find the value of n, we need to determine the value of (n-8)!. We can do this by finding the factorial of numbers from 1 to 8 until we find a number that is equal to or greater than 20160.

By evaluating the factorials, we find that 8! = 40320, which is greater than 20160. The next factorial, 7!, is equal to 5040, which is less than 20160.

Therefore, n must be greater than 8 but less than 16. The only option that satisfies this condition is option 'C', which is 10.

Hence, the correct answer is option 'C'.

A lady arranges a dinner party for 6 guests .The number of ways in which they may be selected from among 10 friends if 2 of the friends will not attend the party together is
  • a)
    164
  • b)
    140
  • c)
    112
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?

Alok Mehta answered
Case I : Both of them don't attend the party.then, no of ways = 8C6 = 28Case 
II : Either of them is selected for party.then, no of ways = 2C1 * 8C5 = 112
Total no of ways = 112+28 = 140

What is the number of diagonals that can be drawn by joining the vertices of a hexagon?
  • a)
    5
  • b)
    12
  • c)
    9
  • d)
    14
Correct answer is option 'C'. Can you explain this answer?

Arun Khanna answered
A diagonal is a line segment joining two non-adjacent vertices of a polygon. A regular hexagon has six sides and six vertices. 

The number of three digit numbers having atleast one digit as 5 is
  • a)
    225
  • b)
    246
  • c)
    648
  • d)
    252
Correct answer is option 'D'. Can you explain this answer?

Ravi Sharma answered
These digit number without digit 5 →100....999
→ these are 900 three-digit number
→ from 100 to 199 → 19 number with 5.
200−299→19
300−399→19
400−499→19
600−699→19
700−799→19
800−899→19
900−999→19
500−599→100
total number with 5=19×8+100 for (500-599)
 =152+100
 =252

 If (n + 1)! = 20(n – 1)!, then n is equal to
  • a)
    20
  • b)
    5
  • c)
    -5
  • d)
    4
Correct answer is option 'D'. Can you explain this answer?

Vikas Kapoor answered
(n + 1)! = 20 (n – 1)!
n (n + 1) = 20
(n – 4) (n + 5) = 0          
Since, (n – 1)! exists, n ≥ 1
So, n = 4 

The number of all three digit even numbers such that if 5 is one of the digits then next digit is 7 is
  • a)
    360
  • b)
    370
  • c)
    365 
  • d)
    none of these
Correct answer is option 'C'. Can you explain this answer?

Raghav Bansal answered
Correct Answer :- c
Explanation : As given condition is , 5 must be followed by 7. So only possible way is 57X where X denotes 0,2,4,6 and 8.
So,total no. of ways=1×1×5=5
Hence, total ways in which we can make a 3-digit even no. without violating given condition are:
360+5=365

The figures 4, 5, 6, 7, 8 are written in every possible order. The number of numbers greater than 56000 is
  • a)
    98
  • b)
    72
  • c)
    90 
  • d)
    none of these
Correct answer is option 'C'. Can you explain this answer?

Riya Banerjee answered
There are in total 6 numbers, 4,5,6,7,8.
Now consider the number 56000
Consider the numbers of the form
56−−−.
Considering no repetitions we get
3×2×1 = 6 numbers.
Similarly for 57−−− and 58−−−.
Hence 3×6 = 18 numbers.
Now consider the numbers starting with 6.
6−−−−
We get 4×3×2×1 = 24.
Similarly for the numbers starting with 7 and 8 we get in total 24 numbers each.
Hence total number of numbers greater than 56000 will be
= (24×3)+18
= 72+18 = 90

In a room there are 2 green chairs, 3 yellow chairs and 4 blue chairs. In how many ways can Raj choose 3 chairs so that at least one yellow chair is included? 
  • a)
    3
  • b)
    30
  • c)
    64
  • d)
    84
Correct answer is option 'C'. Can you explain this answer?

Geetika Shah answered
At least one yellow chair means “total way no yellow chair
Total ways to select 3 chairs form the total (2+3+4) chairs 9C3
9C3 = 9!/3!*6! = 84ways
Now, we dont want even one yellow chair
So, we should select 3 chairs from 6 chairs (2green & 4blue) = 6C3
6C3 = 6!/3!3! = 20 ways
hence , ways to select at least one yellow chair = 84-20 = 64 ways.

On a railway track, there are 20 stations. The number of tickets required in order that it may be possible to book a passenger from every station to every other is
  • a)
    C(20,2)
  • b)
    P(20,2)
  • c)
    400
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?

Milan Shah answered
Solution:

  • There are 20 stations in total.

  • Number of ways to select 2 stations out of 20 is given by C(20,2).

  • So, the number of tickets required to book a passenger from every station to every other station is P(20,2).

  • Therefore, the correct answer is option B.

 What is the value of 0!
  • a)
    zero
  • b)
    1
  • c)
    not defined
  • d)
    2
Correct answer is option 'B'. Can you explain this answer?

Nandini Iyer answered
No zero is not less then zero, i <= 0 becomes 1 because zero is less than or equal to zero. Of course zero isn't less than zero, but i <= 0 becomes 1 because zero is less than or equal to zero.

In a multiple choice question, there are 4 alternatives, of which one or more are correct. The number of ways in which a candidate can attempt this question is
  • a)
    4
  • b)
    25
  • c)
    16
  • d)
    none of these
Correct answer is option 'D'. Can you explain this answer?

Geetika Shah answered
To solve this problem, we need to calculate the total number of ways a candidate can attempt the multiple-choice question where there are 4 alternatives, and one or more can be correct.
Step 1: Consider each alternative
Each of the 4 alternatives can either be:
  1. Selected (included in the answer)
  2. Not selected (excluded from the answer)
So, for each alternative, there are 2 choices (select or not select).
Step 2: Calculate the total number of combinations
If there were no restrictions (i.e., selecting none is allowed), the total number of combinations would be 24=16
Step 3: Subtract the invalid case
Since at least one alternative must be selected (one or more are correct), we subtract the one case where none of the alternatives are selected:
24−1=16−1=15

Find the number of diagonals of an n-sided polygon.
  • a)
    n
  • b)
    (n-1)
  • c)
    n(n-1)/2
  • d)
    n(n-3)/2
Correct answer is option 'D'. Can you explain this answer?

Solution:

To find the number of diagonals of an n-sided polygon, we need to understand what a diagonal is and how many diagonals are present in a polygon.

What is a diagonal?

A diagonal is a line segment that connects two non-adjacent vertices of a polygon.

How many diagonals are present in a polygon?

To count the number of diagonals in a polygon, we need to consider each vertex of the polygon and count the number of diagonals that can be drawn from that vertex.

For example, consider a square.

https://www.edurev.in/ckeditor_assets/pictures/8521/content_diagonal1.png" />

There are four vertices in the square. If we start with the top-left vertex and count the number of diagonals that can be drawn from that vertex, we get two diagonals. Similarly, if we count from the top-right vertex, we get two diagonals. If we count from the bottom-left vertex, we get two diagonals, and if we count from the bottom-right vertex, we get two diagonals. So, in total, there are eight diagonals in a square.

This method works for any polygon. For an n-sided polygon, we have n vertices, and if we count the number of diagonals that can be drawn from each vertex, we get:

- From each vertex, we can draw n-3 diagonals (because we cannot draw a diagonal to the adjacent vertices or to itself).
- Since there are n vertices, the total number of diagonals in an n-sided polygon is:

n(n-3)/2

Therefore, the correct option is (D) n(n-3)/2.

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