Aptitude - Permutation and Combination - Discussion

@Pavz.

Thanks for your explanation.

I have a question, in how many ways can a committee of nine people can be formed from ten men and their wives. If no man is to serve on the committee with his wife?

Please solve this.

But many ones falling problems not finding the details solution of the combination. It might be done more details as like:

7C4 = 7!/4!*(7-4)!
= 7*6*5*4!/4!*3*2*1.
= 7*6*5/3*2*1,
= 35.

And that would be understandable to all.
By the by, May I know the basic differences between combination and permutation.

Among 5 children there are 2siblings.in how many ways children be seated in a row so that the siblings do not sit together.


A) 86
B) 72
C) 46
D) 38
How to solve it? Please anyone explain me.

@Dadasaheb.

The total ways-5!=120.
if siblings sit together-2!4!=48.
Hence; 120-48=72.

I am not getting the solution. Please, anyone explain to me.

Basic Formula is nPr=n!/(n-r)!
n=number
r=spaces

And in case of repetition(if number/character is repeating) we have to apply this formula :nPr/(No .of repeated number/character)! ==>n!/(n-r)!/(No .of repeated number/character)!

Based On Question :

We got three possibilities:

First possibility:- (7 men 6 women) out of which we can select (3 men 2 women)
=>>(7P3/3! * 6P2/2!)
==>(35*15)
{men repeated 3 times and women 2 times that's why i have take 3! and 2!}
==>525----> (1)

Second possibility:-(7 men 6 women) out of which we can select (4 men 1 women)
==>(7P4/4! * 6P1/1!)==>35*6 = 210-----> (2)

Third possibility;-(7 men:6) out of which we can select (5 men only )
==>7P5/5! = 21----> (3)

Add all equatations (1),(2) and (3)
==>525 + 210 + 21 = 756.

Why can't we do this? Select 3 out of 7 men. Because that's the main condition. Then select 2 women out of 6/1 man out of 4 and 1 woman out of 6/ 2 men out of 4.

i.e 7c3 (4c2 + 6c2 + 4c1. 6c1)?

Please tell me.

Here 5 persons are select, and at least 3 men including this committee so we have
Men. Women
7c3 * 6c2 = 525
7c4 * 6c1 = 210
7c5 * 6c0 = 021
= 756 is the answer.

There are 5members required so why don't we divide with 11c5?