Problem :

A committee of 12 is to be formed from nine women and eight men. In how many ways can this be done if at least five women have to be included in a committee ?

Solution :

Case I : 5 W & 7 M = $^9C_5 \times ^8C_7$

Case II : 6W & 6M = $^9C_6 \times ^8C_6$

Case III : 7W & 5 M = $^9C_7 \times ^8C_5$

Case IV : 8W & 4M = $ ^9C_8 \times ^8C_4$

Case V : 9W & 3M = $ ^9C_9 \times ^8C_3$

Adding all the four cases I got the answer.

However I have doubt while doing this problem in other way :

Choosing 5 Women ( at least ) can be done in $^9C_5 \times ^{12}C_7$ $\Rightarrow$ choosing at least five women and choosing rest 7 people out of 12 remaining people.

But I am not getting the correct answer by using this method please suggest the error at this place . Thanks...


Let's call the women A, B, C, D, E, F, G, H, I and the men S, T, U, V, W, X, Y, Z. If you are going to choose $5$ of the women and then $7$ of the rest, here is one way to do it:

(1) choose the women A, B, C, D, E and then choose Z, Y, X, W, V, U, I.

Here is another way:

(2) choose the women A, B, C, D, I and then choose Z, Y, X, W, V, U, E.

But actually the results of these two choices are the same. So you have counted this possibility twice - in fact, if you think about it, more than twice. If you do all the calculations you should find that your second answer is greater than the correct answer.


Reverse the problem: At least 5 women means at most 7 men.

So no 8 men.

Then you can calculate all possibilities, subtract the ones with 8 men.

I guess you can take it from here :)


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