Combination Questions and Answers updated daily – Aptitude


Combination Questions and Answers updated daily – Aptitude



20 Combination Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Combination online test. Combination Questions with detailed description, explanation will help you to master the topic.

Combination Questions

1. How many Permutations of the letters of the word APPLE are there? 



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Correct Ans:60
Explanation:
In the given word “APPLE”, letter "P" is written twice.
So word APPLE contains 1A, 2P, 1L and 1E.
Required Permutations = 5! / 2!
= (5 x 4 x 3 x 2 x 1) / (2 x 1)
= 60


2. Find the sum of all three digit numbers formed by using the digits 1, 2 and 3? 



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Correct Ans:5994
Explanation:
Formula used: sum of integers that are formed by the permutations of n digits
sum is given by equation:
= (sum of digits)*(n-1)!*(111... n times) if repetition is not allowed.
= (sum of digits)*(n)^(n-1)*(111... n times) if repetition is allowed.


As nothing has been mentioned in the question, we'll assume that repetition of numbers are allowed.
= (1 + 2 + 3) * (3)^(3 – 1) * (111)
= 6 * 9 * 111
= 5994


3. Using all the letters of the word "NOKIA", how many words can be formed, which begin with N and end with A? 



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Correct Ans:6
Explanation:
There are five letters in the given word “NOKIA”
Consider 5 blanks ....
The first blank and last blank must be filled with N and A all the remaining three blanks can be filled with the remaining 3 letters in 3! ways.
The number of words = 3! = 3 X 2 X 1 = 6


4. 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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Correct Ans:209
Explanation:
Chances for selecting 4 children From 6 Boys (with atleast 1 boy) From 4 Girls
Chance 1 1 3 -> 6C1 X 4C3
= 6 X 4
= 24
Chance 2 2 2 -> 6C2 X 4C2
= 15 X 6
= 90
Chance 3 3 1 -> 6C3 X 4C1
= 20 X 4
= 80
Chance 4 4 0 -> 6C4
= 15
















Different ways to select 4 children (with atleast 1 boy) = 24 + 90 + 80 + 15 = 209


5. In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together? 



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Correct Ans:720
Explanation:
The vowels in the word 'LEADING' are E, A, I
Number of ways to arrange the letters of the word 'LEADING' such that vowels always come together = LDNG(EAI)
= 5! X 3!
= (5 x 4 x 3 x 2 x 1) X (3 x 2 x 1)
= 120 X 6
= 720


6. In how many ways 6 people can be arranged in row, if one particular person always wants to stand in the rightmost corner? 



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Correct Ans:5!
Explanation:
Out of 6 person one wants to be in the right most corner, so ignore him
Out of the remaining 5 people, they can be arranged in 5! ways.


7. In how many ways a four digit even number can be formed by using the digits 4,5,9,8 exactly once. 



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Correct Ans:12
Explanation:
Given
Four digit even number can be formed by using the digits 4,5,9,8
Since the number has to be a even digit number,the units digit has to be 4 or 8. First three places can be filled by remaining three digits.
Hence it is totally 6 x 2 = 12 ways
Answer is 12


8. What is the sum of two consecutive even numbers, the difference of whose squares is 84? 



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Correct Ans:42
Explanation:
Let the numbers be x and x + 2.
Then, (x + 2)^2 - x^2 = 84
=>4x + 4 = 84
=>4x = 80
=>x = 20.

The required sum = x + (x + 2) = 2x + 2 = 42.


9. 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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Correct Ans:63
Explanation:
Given
Total Men = 7
Total Women = 3
From total, agroup of 5 men and 2 women be made out
Required number of ways
=>(7C5 x 3C2)
=>( 7 * 6 * 5 * 4 * 3 * 2 * 1 ) / (5 * 4 * 3 * 2 * 1) x ( 3 * 2) / ( 2 * 1 )
=>( 7 * 3 ) x ( 3 )
=> 21 * 3
=> 63.


10. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw? 



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Correct Ans:64
Explanation:
We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
= ( 3 x 6 x 5 / 2 x 1 ) + ( 3 x 2 / 2 x 1 x 6) + 1
= (45 + 18 + 1)
= 64


11. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women? 



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Correct Ans:11760
Explanation:
Required number of ways = (8C5 x 10C6)
= (8C3 x 10C4)
= ( 8 x 7 x 6 / 3 x 2 x 1) x (10 x 9 x 8 x 7 / 4 x 3 x 2 x 1)
= 11760


12. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? 



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Correct Ans:25200
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
= (7C3 x 4C2)
= ( 7 x 6 x 5 / 3 x 2 x 1) x (4 x 3 / 2 x 1 )
= 210
Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves = 5! = 5 x 4 x 3 x 2 x 1 = 120.

Therefore, Required number of ways = (210 x 120) = 25200


13. 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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Correct Ans:50400
Explanation:
In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has (6 + 1) = 7 letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters = 7!/2! = 2520.

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5! /3! = 20 ways.

Required number of ways = (2520 x 20) = 50400


14. In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together? 



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Correct Ans:720
Explanation:
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720


15. There are 5 boys and 3 girls in a family. They are photographed in groups of 2 boys and one girl. The number of different photographs will be 



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Correct Ans:180
Explanation:
The number of ways of forming the groups 5C2 x 3C1 = 30
Number of each groups can be arranged among themselves is 3! Ways = 6.
Therefore, the number of photographs = 30 x 6 = 180.


16. Seven points lie on a circle. How many chords can be drawn by joining these points?



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Correct Ans:21
Explanation:
7C2 = 7 x 6 /2 = 21


17. In a hockey championship, there were 153 matches played.Every two teams played one match with each other. The number of teams participating in the championship is 



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Correct Ans:18
Explanation:
Let there were x teams participating in the games, then
Total number of matches = 153
Probablity of matches played
18C2
= 18 x 17 / 2
= 9 x 17
= 153.
So number of teams participated in match =18.


18. A box contains 10 balls out of which 3 are red and the rest are blue. In how many ways can a random sample of 6 balls be drawn from the bag so that at the most 2 red balls are included in the sample and no sample has all the 6 balls of the same color ?



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Correct Ans:168
Explanation:
Given that, Out of 10 balls, 3 balls are red and remaining 7 balls are blue.

The possible ways are as follows:
(i) 1 red ball out of the three and 5 blue balls out of the seven (OR)
(ii) 2 red balls out of the three and 4 blue balls out of the seven

Total number of ways in which a random sample of six balls can be drawn = (3C1 x 7C5) + (3C2 x 7C4)
= [3 * (7*6) / 2] + [3 * (7*6*5) /(3*2)]
= [126 / 2] + [ 630 / 6]
= [63] + [105]
= 168.


19. In an examination paper there are two groups, each containing 4 questions. A candidate is required to attempt 5 questions but not more than 3 questions from any group. In how many ways can 5 questions be selected? 



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Correct Ans:48
Explanation:
A candidate is required to attempt 5 questions but not more than 3 questions from any group.
= 4C3 x 4C2+4C2 x 4C3
= (4 x 6) + (4 x 6)
= 48.


20. What is the remainder left after dividing 1! + 2! + 3! + ....... + 100! by 7?



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Correct Ans:5
Explanation:
From 7! onwards all terms are divisible by 7 as 7 is one of the factor.
So there is no remainder left for those terms i.e. remainder left after dividing 7! + 8! + 9! + ... + 100! is 0.

The only part to be consider is
= 1! + 2! + 3! + 4! + 5! + 6!
= 1 + 2 + 6 + 24 + 120 + 720
= 873

The remainder left after dividing 873 by 7 is 5
Hence, the remainder is 5


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