Permutation and Combination- Aptitude Questions and Answers

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By using digits 1 to 9 how many 4 digit numbers can be formed which has 5 as a unit digit? Repetition of digits is are not allowed.

A)
336

B)
1400

C)
504

D)
251


Correct Answer :

336


Explanation :

5 at the unit place or divisible by 5. So, there is 1 way of doing it.
No repetition of digits.thus there are 8 ways of filling the tens place from the remaining 8 digits (1,2,3,4,6,7,8,9).
the hundreds place can be filled by any of the remaining 7 digits. So, there are 7 ways of filling it.
the thousands place can be filled by any of the remaining 6 digits. So, there are 6 ways of filling it.

Required number of numbers = (1 * 8 * 7 * 6) = 336.

In how many ways can 5 prizes be given away to 7 boys when each boy is eligible for all the prizes?

A)
5!

B)
7!

C)
57

D)
75


Correct Answer :

75


Explanation :

A prize can be given to any one of the 7 boys in (7/1)=7 ways
So, each prize can be distributed in (7/1)=7 ways
5 prizes can be distributed in (7/1)*(7/1)*(7/1)*(7/1)*(7/1) = 75 ways

How many different words, can be formed by rearranging the letters of the word ‘SUCCESS’?

A)
5040

B)
840

C)
420

D)
1680


Correct Answer :

420


Explanation :

The word "SUCCESS" has total 7 elements, there two repeated elements S(thrice) and C(twice). The formula for permutations with repeated elements is as follows, When k out of n elements are indistinguishable, e.g. k copies of the same book, the number of different permutations is n!/k!.

Therefore the total number of arrangement of the word SUCCESS can be formed is 7!/3! *2!
=> (7*6*5*4*3*2*1)/(3*2*1)*(2*1) => 7*6*5*2 => 420

In how many different ways can the letters of the word ‘ARCHIVE’ be arranged?

A)
2630

B)
5040

C)
1680

D)
3456


Correct Answer :

5040


Explanation :

There are 7 unique letters in ARCHIVE which means that there are 7!=7*6*5*4*3*2*1=5040 different ways to rearrange them.

In how many different ways can the letters of the word ‘PUBLIC’ be arranged so that the vowels are at the two ends?

A)
48

B)
72

C)
36

D)
None of these


Correct Answer :

48


Explanation :

PUBLIC has 2 vowels(UI) and 4 consonants (PBLC). All the letters are distinct.
First letter can be any of the two vowels (2 ways) and Last letter is the remaining vowel (1 way)
4 consonants can be arranged in 4! ways

Total number of ways = 2!*1!*4! = (2*1)*1*(4*3*2*1)=48

If 56Pr + 6 : 54Pr + 3 = 30800 : 1 then the value of r is ?

40
41
42
43

Correct Answer :41
Explanation :

56Pr + 6 : 54Pr + 3 = 30800 : 1
⇒ 56! / (56 -(r + 6))! : 54! / (54 - (r + 3))! = 30800 : 1
⇒ 56! / (50 - r)! : 54! / (51 - r)! = 30800 : 1

As 56! = 56*55*54!
and also (51 - r)! =(51 - r)*[(51 - r) -1]! ⇒ (51 - r)*(50 - r)!

So,
⇒ 56*55*54! / (50 - r)! = (30800 x 54!) / (51 - r)*(50 - r)!
⇒ 56*55 = 30800/(51 - r)
⇒ 51 - r = 10
Hence, r = 41

How many groups of 6 persons can be formed from 8 men and 7 women ?

5000
5005
5050
5650

Correct Answer :5005
Explanation :

Total number of person = 8 men + 7 women = 15
No. of groups can be formed of 6 persons= 15C6 = 15! / {6! *(15 - 6)!} = 15! / (6! * 9!)
= (15 x 14 x 13 x 12 x 11 x 10) / (6 x 5 x 4 x 3 x 2 x 1)
= 5005

If 10 persons meet at a meeting and shake hands exactly once with each other, what is the total number of hand shakes?

A)
35

B)
35

C)
50

D)
45


Correct Answer :

45


Explanation :

Combinations of n objects taken r at a time = nCr= n! / [r!(n-r!)]
Required number of ways = 10C2 = 10*9 / 2 = 45

In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is

A)
200

B)
216

C)
235

D)
256


Correct Answer :

200


Explanation :

Let there were g girls and b boys.
Total number of games in which both players were women gC2 = 45
=> g(g−1)/2=45
=> g2-g-90 = 0
(g - 10)(g + 9) = 0
So, g =10 as g ≠ -9

Total number of games in which both players were men bC2 = 190
=>b(b−1)/2=190
=> b2-b-380 = 0
(b - 20)(b + 190) = 0
So, b=20 as b ≠ -190

Total number of games= (g + b)C2 = 30C2 = (30 *29) / (2* 1) = 435

Number of games in which one player is a boy and the other is a girl = 435 - 45 - 190 = 200

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