John and Peter are among the nine players a basketball coach

John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?
A. 1/9
B. 1/6
C. 2/9
D. 5/18
E. 1/3

Probability approach:

1/9 choosing John (J);
1/8 choosing Peter (P);
7/7=1 choosing any (A) for the third player;
6/6=1 choosing any (A) for the fourth player;
5/5=1 choosing any (A) for the fifth player.

But scenario JPAAA can occur in \(\frac{5!}{3!}=20\) # of ways (JAPAA, JAAPA, AAAPJ, ... basically the # of permutations of the letters JPAAA, which is \(\frac{5!}{3!}=20\)).

So \(P=\frac{5!}{3!}*\frac{1}{9}*\frac{1}{8}*1*1*1=\frac{5}{18}\).

Combinatorics approach:

P=favorable outcomes/total # of outcomes --> \(P=\frac{C^2_2*C^3_7}{C^5_9}=\frac{5}{18}\).

\(C^2_2=1\) - # of ways to choose Peter and John out of Peter and John, basically 1 way;
\(C^3_7=35\) - # of ways to choose 3 other players out of 7 players left (without Peter and John);
\(C^5_9=126\) - total # of ways to choose 5 players out of 9.

Answer: D.

Hope it helps.