number of combinations

• Jan 24th 2007, 06:20 AM
Rimas
number of combinations
Given Twenty Couples how many different three member committees can be fored that do not contain both members of any of these couples
• Jan 24th 2007, 07:45 AM
Soroban
Hello, Rimas!

Quote:

Given twenty couples, how many different 3-member committees can be formed
that do not contain both members of any of these couples?

There are 40 people available.

The first member $(A)$ can be any of the 40 people.

Of the remaining 39 people, the second member $(B)$ has 38 choices.
. . (He/she must not be A's spouse.)

Of the remaining 38 people, the third member $(C)$ has 36 choices.
. . (He/she must not be A's spouse or B's spouse.)

Hence, there are: . $40 \times 38 \times 36 \:=\:54,720$ ways.

But these selections include an ordering of the members.
That is, $\{A,B,C\}$ is considered a different committe from $\{B,C,A\}$

Since three members can be ordered in $3! = 6$ ways,
. . we divide by 6 to eliminate the duplications.

Answer: . $\frac{54,720}{6}\:=\:9,120$ ways.