+0

# Prob

0
76
3
+278

Suppose that the Senate has 47 Republicans and 53 Democrats. In how many ways can we form a 3-senator committee in which neither party holds all 3 seats?

My work so far: We can start by choosing from any. So thats 100. Then, we can still choose from any remaining, thats 99. Then, we can choose from either one of the republicans if we started with democrats, or democrats if we started with republicans. Please help, how do I continue.

Mar 24, 2020

### 3+0 Answers

#1
+23710
+2

Here is my shot at this:

You can choose 2 democrats and one repub   or    two repub and 1 dem.

47 c 2   x  53      +    53 c 2 x 47   =......

Mar 24, 2020
#2
+485
+2

My advice would be to use complementary counting. In otherwords, you take the total number of ways to form a 3 senator committee, and subtract from it the number of ways that a single party can hold all 3 seats. That simplifies the problem immensely.

We then have:

100 * 99 * 98 ways to form a 3 member senate. If we had a committe with all 3 seats belonging to one party, there are two cases:

Democrats occupy all of them, of which there are:

47*46*45 ways to do this

Republicans occupy all of them, of which there are:

53*52*51

We add these two cases up, and then subtract it from the total number of ways to form a 3 member senate.

970200- 237846= 732354 ways(check me if I'm wrong)

Mar 24, 2020
edited by jfan17  Mar 24, 2020
#3
+278
+2

Thank you for the hints and help guys!!!!

Mar 24, 2020