Permutation and Combination- Aptitude Questions and Answers - in Hindi

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

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.

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

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.

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

⁵⁶P_{r + 6} : ⁵⁴P_{r + 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

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

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

Let there were g girls and b boys.
As given, 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

As given, 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

DESIGN has 2 vowels(EI) and 4 consonants (DSGN).
Four consonants can be arranged in 4! ways two vowels can be arranged in 2! ways

Total number of arrangements= 4!×2! = 48