Question 890885
A United States Delegation consists of 6 Americans, 5 Russians, and 4 Chinese.
How many committees of three people have more Americans than Russians?
<pre>
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.
</pre>
AND

How many committees of three people do not have all three Americans
<pre>
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</pre>