Question

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

Answer 1

I think it is\[\left[\binom{7}{3}\times \binom{6}{2}\right] + \left[\binom{7}{4} \times \binom{6}{1}\right] + \left[\binom{7}{5}\right]\]

Answer 2

Since the possible ways are:

3 men and 2 women
4 men and 1 woman
5 men

Answer 3

I prefer
\[\Large \left(\begin{matrix}7\\ 3\end{matrix}\right)\times \left(\begin{matrix}10\\ 2\end{matrix}\right)\]

3 men and 2 more of heaven-knows-what-gender

Answer 4

Whom do i prefer? :P

Answer 5

I don't know... :D

Answer 6

ohk.. @terenzreignz

Answer 7

Well... @ParthKohli ?

Answer 8

Hmm, our answers are different. But I'm not able to get who has their answer incorrect lol

Answer 9

ehh.. stick to @ParthKohli
I probably oversimplified to ... oblivion :D

Answer 10

Today is not my day o.O

Answer 11

It's not mine either.

Answer 12

@terenzreignz your answer is def not correct :P

Answer 13

That settles it.

Answer 14

Hmm, @terenzreignz now the more i think about it, the more I wonder why your answer doesn't work... it seems reasonable, but gives over double the possible combinations as parth's method.

Yours guarantees having at least 3 men... but I'm guessing it counts a few situations twice. ie you pick Joe, Barry and Larry from the first 7, then Harry and Sam from the second group of 10.

Then you could also pick Joe, sam and harry from the first 7, then also pick Barry and Larry from the second group of 10, etc.

Answer 15

I've already put it out of my mind, unfortunately :D
But Parth's method seems so fail-safe...so

Answer 16

His method avoids possibly over-counting the same group twice, or thrice, or... four... ice? Frice?

Answer 17

Possible groupings :-
1. 3 men and 2 women
2. 4 men and 1 woman
3. 5 men

These 3 results are independent of each other, hence we got to calculate each individually and add the 3 results.

No of ways to choose 3 men from 7 and 2 women from 6 = 7C3 x 6C2=525
No of ways to choose 2 men from 7 and 1 women from 6 = 7C2 x 6C1=126
No of ways to choose 5 men from 7 and 0 women from 6 = 7C5 x 6C0=21
Total no. of ways= 525+126+21= 672
(This is a problem on Combinations) I hope you are familiar with Permutations and Combinations?

Answer 18

No of ways to choose 2 men from 7 and 1 women from 6 = 7C2 x 6C1=126

@anikhalder that should be choosing 4 men.

Answer 19

oh yea..i'm sorry..it would be 4..thanks @agent0smith

and 1 woman instead of 1 women..:)

7C4 x 6C1= 210
So total is= 525+210+21=756