A United States Delegation consists of 6 Americans, 5 Russians, and 4 Chinese.
How many committees of three people have more Americans than Russians?
Case 1: 1 American, 0 Russians, 2 Chinese
6 Americans choose 1, C(6,1) = 6
5 Russians choose 0, C(5,0) = 1
4 Chinese choose 2, C(4,2) = 6
6 = 36 ways
Case 2: 2 Americans, 0 Russians, 1 Chinese
6 Americans choose 2, C(6,2) = 15
5 Russians choose 0, C(5,0) = 1
4 Chinese choose 1, C(4,1) = 4
15 = 60 ways
Case 3: 2 Americans, 1 Russian, 0 Chinese
6 Americans choose 2, C(6,2) = 15
5 Russians choose 1, C(5,1) = 5
4 Chinese choose 0, C(4,0) = 1
15 = 75 ways
Case 4: 3 Americans, 0 Russians, 0 Chinese
3 Americans choose 2, C(6,3) = 20
0 Russians choose 1, C(5,1) = 1
0 Chinese choose 0, C(4,0) = 1
20 = 20 ways
Answer: 36+60+75+20 = 191 committees with more Ameericans than Russaians.
AND
How many committees of three people do not have all three Americans
First we calculate the number of possible committees of three people.
There are 6+5+4 = 15 people
15 people choose 3 = C(15,3) = 455 committees
From that we subtract the number of all American committees.
6 Americans choose 3 = C(6,3) = 20
Answer = 455 - 20 = 435
Edwin
