SOLUTION: a) In how many ways can a committee consisting of three men and two women be chosen from seven men and five women? b) How many committee of five with a given chairman can be sel

We can choose the 3 men C(7,3) ways. 
For each of those ways, we can choose the 2 women C(5,2) ways.

Answer C(7,3)C(5,2) = 35*10 = 350 ways

I can't tell for sure whether you mean that we must select only one particular
man, say John Smith, as the only person who can be the chairman, and none of the
other 11 can be the chairman. 

or whether 

Any of the 12 men can be the chairman.

So I'll do it both ways:

A.  If it's the first way (which I doubt), then 

we choose theit chairman 1 way, C(1,1), as John Smith,
then we choose the 4 non-chairmen C(11,4) ways.

Answer: if it's that way:  C(1,1)*C(11,4) = 1*330 = 330 ways.
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B.  If it's the second way (which I suspect),   

There are two correct ways to approach the problem:  Choose the chairman first,
and then choose the 4 non-chairmen, OR choose the committee first, then choose
the chairman.

   1.  We can pick the chairman 12 ways.  C(12,1)
       For each of those ways we can then choose the 4 non-chairmen C(11,4)
       ways.
       Answer: C(12,1)C(11,4) = 12*330 = 3960 ways.  

   2.  We can pick the committee C(12,5) ways.
       For each way to pick a committee we can pick a chairman 5 ways. C5,1)
       Answer: C(12,5)C(5,1) = 792*5 = 3960 ways.

Take your pick of those two.  They are both correct. Notice that although they
multiply different numbers they both end up giving the same answer 3960.
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Let the 6 students be A,B,C,D,E, and F.

The teacher can make 2 choices about A.  They are: 
1. Choose A  
2. Don't choose A
That's 2 choices the teacher can make about A.

For each of those 2 choices the teacher can make about A,
the teacher can make 2 choices about B.  They are: 
1. Choose B  
2. Don't choose B.
That's 2*2=22 = 4 choices the teacher can make about A and B.

For each of those 4 choices the teacher can make about A and B,
the teacher can make 2 choices about C.  They are: 
1. Choose C  
2. Don't choose C.
That's 2*2*2=23 = 8 choices the teacher can make about A,B and C.

For each of those 8 choices the teacher can make about A,B and C,
the teacher can make 2 choices about D.  They are: 
1. Choose D  
2. Don't choose D.
That's 2*2*2*2=24 = 16 choices the teacher can make about A,B,C, and D.

For each of those 16 choices the teacher can make about A,B,C and D,
the teacher can make 2 choices about E.  They are: 
1. Choose E  
2. Don't choose E.
That's 2*2*2*2*2=25 = 32 choices the teacher can make about A,B,C,D, and E.

For each of those 32 choices the teacher can make about A,B,C,D, and E,
the teacher can make 2 choices about F.  They are: 
1. Choose F  
2. Don't choose F.
That's 2*2*2*2*2*2=26 = 64 choices the teacher can make about A,B,C,D,E and F.

However, among those 64 possible choices the teacher can make, we must
eliminate the one case where the teacher does not choose any of the 6 
students.  So we must subtract that 1 case from the 64.

Answer: 64-1 = 63 

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If the 3 U's looked different (were distinguishable like this
UNUSUAL, the answer would be 7! 

However, since we cannot tell the difference between the U's, we
must divide by the number of positions the three U's can be placed
within each permutation, which is 3! so the answer is  ways.

Edwin