Given Twenty Couples how many different three member committees can be fored that do not contain both members of any of these couples
Hello, Rimas!
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 $\displaystyle (A)$ can be any of the 40 people.
Of the remaining 39 people, the second member $\displaystyle (B)$ has 38 choices.
. . (He/she must not be A's spouse.)
Of the remaining 38 people, the third member $\displaystyle (C)$ has 36 choices.
. . (He/she must not be A's spouse or B's spouse.)
Hence, there are: .$\displaystyle 40 \times 38 \times 36 \:=\:54,720$ ways.
But these selections include an ordering of the members.
That is, $\displaystyle \{A,B,C\}$ is considered a different committe from $\displaystyle \{B,C,A\}$
Since three members can be ordered in $\displaystyle 3! = 6$ ways,
. . we divide by 6 to eliminate the duplications.
Answer: .$\displaystyle \frac{54,720}{6}\:=\:9,120$ ways.