How may comissions can be formed (details below)?

Question

In a company there are 100 employees, 50 men and 50 women. Out of the 50 men, 18 are fluent in French and don't know Math, 17 have a Phd in Math but don't speak French and 7 are fluent in French and have a Phd in Math.

Out of the women, 21 are fluent in French but don't know Math, 13 have a Phd in Math but don't speak French and 10 are fluent in French and have a Phd in Math.

The CEO wants to form a comission of 20 people, according to the following rules:

There must be 10 men and 10 women.
At least 8 people must be fluent in French
At least 9 people must have a Phd in Math.

How many comissions can the CEO form?

Thank you

Answer 1

I think I'm interpreting this question differently from ksoileau, above. First of all there are 50C10 ways of choosing 10 men to form the committee, thus (50C10)^2 ways of choosing the committee. However, not all of these will meet the french and math requirements. But this is just to give an idea of the magnitude of the numbers we're working with.

Let's divide the 100 people into 8 categories.
MEN:
A. French only, 18
B. Math only, 17
C. Both, 7
D. Neither, 8

WOMEN:
E. French only, 21
F. Math only, 13
G. Both, 10
H. Neither, 6

What greatly complicates this problem is that there are so many possible cases to look at. Take first of all the cases where the women meet all the french and math requirements for the committee.

Case 1: 10 from G
Then there are 50C10 ways of choosing the men

Case 2: 9 from G, 1 other woman
There are 10 ways of choosing 9 from G, 40 ways of choosing the other woman, and 50C10 ways of choosing the men

Case 3: 8 from G, 1 from F, 1 other woman
There are 10C8 ways of choosing 8 from G, 13 ways of choosing 1 from F, 27C1 ways of choosing the other woman, and 50C10 ways of choosing the men

Case 4: 7 from G, 2 from F, 1 from E
There are 10C7 ways of choosing 7 from G, 13C2 ways of choosing 2 from F, 21 ways of choosing 1 from E, in addition to 50C10 ways of choosing the men

So far there are 50C10 * (1+10*40+ 45*13*27+120*78*21) = 212756*(50C10) ways this committee can be chosen with the women meeting all the requirements.

There is one case where the men meet all the requirements.
Case 5: 7 from C, 2 from B, 1 from A
There are 17C2 ways of choosing 2 from B, 18 ways of choosing 1 from A. There are also (50C10 - 212756) ways of choosing the women, which we have not already counted.
So that's an additional 4896 * (50C10 - 212756).

Then you have to look at the combinations where neither gender meets the requirements individually, but together.

But already I've counted 2.235 x 10^15 combinations. Maybe I'm completely missing the mark, but this is how I see it.

Answer 2

Let a committee be represented by the values of the following variables:
a=# of males on committee fluent in French, no PhD
b=# of males on committee with PhD, no French
c=# of males on committee fluent in French, with PhD
d=# of males on committee with no French, no PhD
e=# of females on committee fluent in French, no PhD
f=# of females on committee with PhD, no French
g=# of females on committee fluent in French, with PhD
h=# of females on committee with no French, no PhD

Then a committee with composition (a,b,c,d,e,f,g,h)
will have a+c+e+g French speakers, e+f+g+h females, a+b+c+d males and b+c+f+g PhD's.
To satisfy the constraints of an acceptable committee,
the following constraints must be satisfied:
a+c+e+g>=8
e+f+g+h=10
a+b+c+d=10
b+c+f+g>=9
0<=a<=18
0<=b<=17
0<=c<=7
0<=d<=8
0<=e<=21
0<=f<=13
0<=g<=10
0<=h<=6

OK, I get 4453 different committees. I'll have to check my work...

Here's ten of them
(5,2,1,2,1,4,2,3)
(5,2,1,2,1,4,3,2)
(5,2,1,2,1,4,4,1)
(5,2,1,2,1,4,5,0)
(5,2,1,2,1,5,1,3)
(5,2,1,2,1,5,2,2)
(5,2,1,2,1,5,3,1)
(5,2,1,2,1,5,4,0)
(5,2,1,2,1,6,1,2)
(5,2,1,2,1,6,2,1)