Permutation and Combination1
Permutation and Combination1
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Question 1 of 30
1. Question
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
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Question 2 of 30
2. Question
In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
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Question 3 of 30
3. Question
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
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Question 4 of 30
4. Question
In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?
CorrectIncorrectHint
The word ‘OPTICAL’ has 7 letters. It has the vowels ‘O’,’I’,’A’ in it and these 3 vowels
should always come together. Hence these three vowels can be grouped and considered as a
single letter. That is, PTCL(OIA).
Hence we can assume total letters as 5. and all these letters are different.
Number of ways to arrange these letters = 5! = [5 x 4 x 3 x 2 x 1] = 120
All The 3 vowels (OIA) are different
Number of ways to arrange these vowels among themselves = 3! = [3 x 2 x 1] = 6
Hence, required number of ways = 120 x 6 = 720 
Question 5 of 30
5. Question
In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?
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Question 6 of 30
6. Question
In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
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Question 7 of 30
7. Question
In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged such that the vowels must always come together?
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Question 8 of 30
8. Question
There are 8 men and 10 women and you need to form a committee of 5 men and 6 women. In how many ways can the committee be formed?
CorrectIncorrect 
Question 9 of 30
9. Question
How many 3letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?
CorrectIncorrectHint
The word ‘LOGARITHMS’ has 10 different letters.
Hence, the number of 3letter words(with or without meaning) formed by using these letters
= ^{10}P_{3}
= 10 x 9 x 8
= 720 
Question 10 of 30
10. Question
In how many different ways can the letters of the word ‘LEADING’ be arranged such that the vowels should always come together?
CorrectIncorrectHint
The word ‘LEADING’ has 7 letters. It has the vowels ‘E’,’A’,’I’ in it and
these 3 vowels should always come together. Hence these 3 vowels can be grouped
and considered as a single letter. that is, LDNG(EAI).
Hence we can assume total letters as 5 and all these letters are different.
Number of ways to arrange these letters = 5! = 5 x 4 x 3 x 2 x 1 = 120
In the 3 vowels (EAI), all the vowels are different.
Number of ways to arrange these vowels among themselves = 3! = 3 x 2 x 1= 6
Hence, required number of ways = 120 x 6= 720 
Question 11 of 30
11. Question
A coin is tossed 3 times. Find out the number of possible outcomes.
CorrectIncorrectHint
When a coin is tossed once, there are two possible outcomes – Head(H) and Tale(T)
Hence, when a coin is tossed 3 times, the number of possible outcomes
= 2 x 2 x 2 = 8
(The possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT ) 
Question 12 of 30
12. Question
In how many different ways can the letters of the word ‘DETAIL’ be arranged such that the vowels must occupy only the odd positions?
CorrectIncorrectHint
The word ‘DETAIL’ has 6 letters which has 3 vowels (EAI) and 3 consonants(DTL)
The 3 vowels(EAI) must occupy only the odd positions.
Let’s mark the positions as (1) (2) (3) (4) (5) (6).
Now, the 3 vowels should only occupy the 3 positions marked as (1),(3) and (5) in any order.
Hence, number of ways to arrange these vowels = ^{3}P_{3}
= 3! = 3 x 2 x 1 = 6
Now we have 3 consonants(DTL) which can be arranged in the remaining 3 positions in any order
Hence, number of ways to arrange these consonants = ^{3}P_{3}
= 3! = 3 x 2 x 1 = 6
Total number of ways
= number of ways to arrange the vowels x number of ways to arrange the consonants
= 6 x 6 = 36 
Question 13 of 30
13. Question
A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw?
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Question 14 of 30
14. Question
In how many different ways can the letters of the word ‘JUDGE’ be arranged such that the vowels always come together?
CorrectIncorrectHint
The word ‘JUDGE’ has 5 letters. It has 2 vowels (UE) in it and these 2 vowels should
always come together. Hence these 2 vowels can be grouped and considered as a single
letter. That is, JDG(UE).
Hence we can assume total letters as 4 and all these letters are different.
Number of ways to arrange these letters = 4!= 4 x 3 x 2 x 1 = 24
In the 2 vowels (UE), all the vowels are different.
Number of ways to arrange these vowels among themselves = 2! = 2 x 1 = 2
Total number of ways = 24 x 2 = 48 
Question 15 of 30
15. Question
In how many ways can the letters of the word ‘LEADER’ be arranged?
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Question 16 of 30
16. Question
How many arrangements can be made out of the letters of the word ‘ENGINEERING’ ?
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Question 17 of 30
17. Question
How many 3 digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated?
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Question 18 of 30
18. Question
How many words with or without meaning, can be formed by using all the letters of the word, ‘DELHI’ using each letter exactly once?
CorrectIncorrectHint
The word ‘DELHI’ has 5 letters and all these letters are different.
Total words (with or without meaning) formed by using all these
5 letters using each letter exactly once
= Number of arrangements of 5 letters taken all at a time
= ^{5}P_{5} = 5! = 5 x 4 x 3 x 2 x 1 = 120 
Question 19 of 30
19. Question
What is the value of ^{100}P_{2} ?
CorrectIncorrectHint
^{100}P_{2} = 100 x 99 = 9900

Question 20 of 30
20. Question
In how many different ways can the letters of the word ‘RUMOUR’ be arranged?
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Question 21 of 30
21. Question
There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period?
CorrectIncorrectHint
We have 6 periods and need to organize 5 subjects such that each subject is allowed
at least one period.
In 6 periods, 5 can be organized in ^{6}P_{5} ways.
Remaining 1 period can be organized in ^{5}P_{1} ways.
Total number of arrangements
= ^{6}P_{5} x ^{5}P_{1}
= (6 x 5 x 4 x 3 x 2 ) x (5)
= 720 x 5
= 3600 
Question 22 of 30
22. Question
How many 6 digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once?
CorrectIncorrectHint
The first two places can only be filled by 3 and 5 respectively and there is only 1 way
of doing this
Given that no digit appears more than once. Hence we have 8 digits remaining(0,1,2,4,6,7,8,9)
So, the next 4 places can be filled with the remaining 8 digits in ^{8}P_{4} ways
Total number of ways = ^{8}P_{4} = 8 x 7 x 6 x 5 = 1680 
Question 23 of 30
23. Question
An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of table and chair?
CorrectIncorrectHint
He has has 10 patterns of chairs and 8 patterns of tables
Hence, A chair can be arranged in 10 ways and
A table can be arranged in 8 ways
Hence one chair and one table can be arranged in 10 x 8 ways = 80 ways 
Question 24 of 30
24. Question
25 buses are running between two places P and Q. In how many ways can a person go from P to Q and return by a different bus?
CorrectIncorrectHint
He can go in any bus out of the 25 buses.
Hence He can go in 25 ways.
Since he can not come back in the same bus that he used for travelling,
He can return in 24 ways.
Total number of ways = 25 x 24 = 600 
Question 25 of 30
25. Question
A box contains 4 red, 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours?
CorrectIncorrectHint
1 red ball can be selected in ^{4}C_{1} ways
1 white ball can be selected in ^{3}C_{1} ways
1 blue ball can be selected in ^{2}C_{1} ways
Total number of ways
= ^{4}C_{1} x ^{3}C_{1} x ^{2}C_{1}
=4 x 3 x 2
= 24 
Question 26 of 30
26. Question
A question paper has two parts P and Q, each containing 10 questions. If a student needs to choose 8 from part P and 4 from part Q, in how many ways can he do that?
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Question 27 of 30
27. Question
In how many different ways can 5 girls and 5 boys form a circle such that the boys and the girls alternate?
CorrectIncorrectHint
In a circle, 5 boys can be arranged in 4! ways
Given that the boys and the girls alternate.
Hence there are 5 places for girls which can be arranged in 5! ways
Total number of ways = 4! x 5! = 24 x 120 = 2880 
Question 28 of 30
28. Question
Find out the number of ways in which 6 rings of different types can be worn in 3 fingers?
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Question 29 of 30
29. Question
In how many ways can 5 man draw water from 5 taps if no tap can be used more than once?
CorrectIncorrectHint
1^{st} man can draw water from any of the 5 taps
2^{nd} man can draw water from any of the remaining 4 taps
3^{rd} man can draw water from any of the remaining 3 taps
4^{th} man can draw water from any of the remaining 2 taps
5^{th} man can draw water from remaining 1 tap5 4 3 2 1 Hence total number of ways = 5 x 4 x 3 x 2 x 1 = 120

Question 30 of 30
30. Question
In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?
CorrectIncorrect