Aptitude - Permutation and Combination - Discussion

First of all we need to select 3 man from 7 so = 7C3 = 35
Then we can choose 2 person from (7-3)+ 6 person ie. 10C2 = 45
So answer should be 53*45 = none of these..

Please tell me where I'm wrong.

Hi Naveen,

you are wrong

first step is correct that is
First of all we need to select 3 man from 7 so = 7C3 = 35
and second step you are wrong

in problem they are asking minimum of 3 men so already you took 3 men
so now we need woman

so we need 2 women from 6 women

so 6c2
so the value of 6c2 is 15

now 35*15 = 525

like same way second committee we took 4 men from 7 men group
i.e. 7c4 = 35

we need one more woman for 2nd committe because already we took 4 men

so 6c1 = 6

so 35*6 = 210

now we took all men for 3 rd commitee so 7c5 = 21

now the total is 525+210+21 = 756

I hope you have understood the problem.

Nice answer nagu.

Why would the logic 7C3 * 10C2 not work? First we've choosen 3 Men - 7C3. Now we have 10 people left (which included both men and women). Then we need to select any 2 people either men or women so 10C2. Please Clarify.

@Nikhil,

The combinations in 7C3 and 10C2 are not mutually exclusive. Hence we cannot just multiply both and result in a solution.

Its like saying, the combinations that 7C3 gives is same as what
7C1*6C1*5C1 would give.(selecting 1 out of 7 as first Man, 1 out of remaining 6 men for second...etc)

That is 35 is not equal to 210.

In Above solution I am unable to understand why have we included 5/1(in both the committee and 6/2 in 2nd committee. Can any help me to understand how the combination considered in the example.

5 ball are to be placed in 12 boxes, they are placed in three rows such that each row contain at least one ball in how many ways it can be placed.

Not a good answer to give please another method please easy method.

How can you take a 3 men 2 women 4men and 1women and a 5 men please solve easy method.

In the first step 7c5 converted 7c2 in the second step. But how ?