1. There are 2 senators from each of the 50 states.
We wish to make a 3-senator committee in which no two of the members are from the same state. In how many ways can we do it?
2. Suppose for this problem (though it may not be accurate in real life) 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?
thx
+31 1. For the First Senator, we have 100 ways to choose them.
For the Second Senator, we have 98 senators to choose from(Since we already have 1 senator and we don't want to choose the senator from the same state as the first)
For the Third Senator, we have 96 Senators to choose from(This is using the same logic.)
So, we have 100*98*96 Senators, but we want to know the amount without ordering, so we divide it by 6. This equates to 156800
2. For this problem, we can find all the ways You can form a committee of 3 senators and subtract the ways in which a party holds all 3 seats.
The total amount of ways You can form a committee of 3 senators is 100 choose 3 which is 161700.
The amount of ways to choose 3 Democrats from 53 Democrats is 53 Choose 3, which is 23426
The Amount of ways to choose 3 Republicans from 47 Republicans is 47 Choose 3, which is 16215
Therefore, the amount of ways is 161700-23426-16215 = 122059
