Aptitude - Permutation and Combination - Discussion

Yes it's write answer.

Using combination formula.
nCr = n! / r! ( n - r ) !
Question says at least 3 men's , means we have to select minimum 3 men's , it means we can select either 3 men's or 4 men's or 5 men's.
1)in first case select 3 men's and 2 women's. To form 5 people commette ..
7! / 3! (7-3)! * 6! / 2!(6-2)! - Eq-1
2 ) in second case select 4 men's and 1 women can be selected in.
7! / 4!(7-4)! * 6! / 1!(6-1)! - Eq -2
3) in third cas we can select 5 men's only
Can be selected in 7! / 5!(7-5)! - Eq -3

The final answer is:

Eq-1 + Eq-2 + Eq-3=
7! / 3! (7-3)! * 6! / 2!(6-2)! + 7! / 4!(7-4)! * 6! / 1!(6-1)! + 7! / 5!(7-5)!
= 525 + 210 + 21
=756 ways.

Why can't we do it like this,

Total ways = at least 3 men + not men.
i.e not men = 6c5.
Total ways= 13c5.
therefore,
13c5 - 6c5.

@Dadasaheb.

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

@Nivetha.

We only have to find total combinations, not the probability. If we want to find probability then we have to divide it by 13C5.

nCr = nC(n-r) --> formula.

7C3 = 7C(7-3) = 7C4 = 35.

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.

@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.

How many 3 digit numbers can be formed from 2, 3, 4, 5, 6, 7, 9, which are divisible by 5 and none of the digits is repeated? Please tell answer and method for this problem.

Remarked first line 7c5 second line 7c2. How?

I did not understand how we got 21 for 7C5?