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- Most Important Multiple Choice Questions
- Online Permutation and Combination Exercise with Correct Answer Key and Solutions
- Useful for all Competitive Exams
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- Question 1 of 22
1. Question
In how many different ways can the letters of the word BOOKLET be arranged such that B and T always come together?
Hint
Treat B and T as a single letter. Then the remaining letters (5 + 1 = 6) can be arranged in 6! ways. Since, O is repeated twice, we have to divide by 2 and the B and T letters can be arranged in 2! ways.
Total no. of ways
- Question 2 of 22
2. Question
In how many different ways can the letters of the word RUMOUR be arranged?
Hint
Required number of ways
- Question 3 of 22
3. Question
In how many different ways can the letters of the word JUDGE be arranged so that the vowels always come together?
Hint
Required number
= 4! × 2! = 24× 2 = 48
- Question 4 of 22
4. Question
How many words can be formed from the letters of the word SIGNATURE so that the vowels always come together?
Hint
The word SIGNATURE consists of nine letters comprising four vowels (A, E, I and U) and five consonants (G, N, R, T and S). When the four vowels are considered as one letter, we have six letters which can be arranged in ⁶P₆ ways ie 6! ways. Note that the four vowels can be arranged in 4! ways.
Hence required number of words = 6! × 4! = 720 × 24 = 17280
- Question 5 of 22
5. Question
If all permutations of the letters of the word AGAIN are arranged as in dictionary, then fiftieth word is
Hint
Starting with the letter A, and arranging the other four letters, there are 4! = 24 words.
These are the first 24 words.
Then starting with G, and arranging A, A, I, and N in different ways, there are
words.Hence, total 36 words.
Next, the 37th word starts with I.
There are 12 words starting with I.
This accounts up to the 48th word.
The 49th word is NAAGI. The 50th word is NAAIG.
- Question 6 of 22
6. Question
All the words that can be formed using alphabets A, H, L, U and R are written as in a dictionary (no alphabet is repeated). Rank of the word RAHUL is
Hint
No. of words starting with A are 4 ! = 24
No. of words starting with H are 4 ! = 24
No. of words starting with L are 4 ! = 24
These account for 72 words
Next word is rahlu and the 74th word rahul.
- Question 7 of 22
7. Question
How many ways are there to arrange the letters in the word GARDEN with vowels in alphabetical order
Hint
Total number of arrangements of letters in the word GARDEN = 6 ! = 720.
There are two vowels A and E, in half of the arrangements A precedes E and other half A follows E.
So, vowels in alphabetical order in
- Question 8 of 22
8. Question
In how many ways a committee consisting of 5 men and 6 women can be formed from 8 men and 10 women?
Hint
Here, 5 men out of 8 men and 6 women out of 10 women can be chosen in
⁸C₅ × ¹⁰C₆ ways
i.e., 11760 ways.
- Question 9 of 22
9. Question
There are 10 lamps in a hall. Each of them can be switched on independently. The number of ways in which the hall can be illuminated is
Hint
Since each bulb has two choices, either switched on or off.
Therefore required number = 2¹⁰ – 1 = 1023.
- Question 10 of 22
10. Question
The number of ways in which 52 cards can be divided into 4 sets, three of them having 17 cards each and the fourth one having just one card
Hint
Here we have to divide 52 cards into 4 sets, three of them having 17 cards each and the fourth one having just one card.
First we divide 52 cards into two groups of 1 card and 51 cards.
This can be done in
ways.Now every group of 51 cards can be divided into 3 groups of 17 each in
.Hence the required number of ways
=
- Question 11 of 22
11. Question
Three dice are rolled. The number of possible outcomes in which at least one die shows 5 is
Hint
Required number of possible outcomes
= Total number of possible outcomes – Number of possible outcomes in which 5 does
not appear on any dice. (hence 5 possibilities in each throw)
= 6³ – 5³ = 216 – 125 = 91
- Question 12 of 22
12. Question
The number of ways in which ten candidates A₁, A₂, …., A₁₀ can be ranked so that A₁ is always above A₂ is
Hint
Ten candidates can be ranked in 10! ways.
In half of these ways A₁ is above A₂ and in another half A₂ is above A₁.
So, required number of ways is
. - Question 13 of 22
13. Question
The number of ways in which n distinct objects can be put into two different boxes is
Hint
Let the two boxes be B₁ and B₂.
There are two choices for each of the n objects.
So, the total number of ways is
. - Question 14 of 22
14. Question
If 12 persons are seated in a row, the number of ways of selecting 3 persons from them, so that no two of them are seated next to each other is
Hint
The number of ways of selecting 3 persons from 12 people under the given condition :
Number of ways of arranging 3 people among 9 people seated in a row, so that no two of them are consecutive
= Number of ways of choosing 3 places out of the 10 [8 in between and 2 extremes]
=
- Question 15 of 22
15. Question
Ten different letters of an alphabet are given, words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated is
Hint
Total number of words that can be formed = 10⁵.
Number of words in which no letter is repeated = ¹⁰P₅.
So, number of words in which at least one letter is repeated
= 10⁵ – ¹⁰P₅ = 69760.
- Question 16 of 22
16. Question
A five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done is
Hint
If a number is divisible by 3, the sum of the digits in it must be a multiple of 3.
The sum of the given six numerals is 0 + 1 + 2 + 3 + 4 + 5 = 15.
So to make a five digit number divisible by 3 we can either exclude 0 or 3.
If 0 is left out, then 5! = 120 number of ways are possible.
If 3 is left out, then the number of ways of making a five digit numbers is 4 × 4! = 96, because 0 cannot be placed in the first place from left, as it will give a number of four digits.
Hence, total number = 120 + 96 = 216.
- Question 17 of 22
17. Question
The number of possible outcomes in a throw of n ordinary dice in which at least one of the dice shows an odd number is
Hint
Total number of ways
= 6 × 6 × ….. to n times = 6n.
Total number of ways to show only
even number = 3 × 3 × …… to n times = 3n.
∴ required number of ways = 6n – 3n.
- Question 18 of 22
18. Question
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at-least one black ball is to be included in the draw?
Hint
At least one black ball can be drawn in the following ways:
(i) one black and two other colour balls
(ii) two black and one other colour balls, and
(iii) all the three black balls
Therefore the required number of ways is
- Question 19 of 22
19. Question
To fill a number of vacancies, an employer must hire 3 programmers from among 6 applicants, and 2 managers from among 4 applicants. What is the total number of ways in which she can make her selection?
Hint
Required no. of the ways
= ⁶C₃ × ⁴C₂ = 20 × 6 = 120
- Question 20 of 22
20. Question
The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by
Hint
No. of ways in which 6 men can be arranged at a round table = (6 – 1)!
Now women can be arranged in 6! ways.
Total Number of ways = 6! × 5!
- Question 21 of 22
21. Question
The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is
Hint
Required number of ways
= coefficient of x² in (x + x² + … x⁶)³
[Since each box can receive minimum 1 and maximum 6 balls]
= coefficient of x⁸ in x²(x + x + x² + … + x⁵)³
= coefficient of x⁵ in
= coefficient of x⁵ in (1 – x)-³
= coefficient of
= ⁷C₅ = 21
- Question 22 of 22
22. Question
From amongst 36 teachers in a school, one principal and one vice-principal are to be appointed. In how many ways can this be done?
Hint
Principal can be appointed in 36 ways.
Vice principal can be appointed in the remaining 35 ways.
Total number of ways = 36 × 35 = 1260
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