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Question : A team of three children plus a captain is required for a school chess competition. A particular school has ten children who can play chess of whom only three are qualified to act as captain.
In how many ways can a team be selected assuming that the three children qualified as captain are also eligible to play chess?

My Try: $(^3C_1\cdot\; ^7C_3) + (^3C_2 \cdot \; ^7C_2) + (\;^3C_3 \cdot \; ^7C_1) = 175$

Claimed Answer : The answer is $252$, however, I was not able to reach here. If Anyone know how to solve this question, please guide.

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  • $\begingroup$ What answer are you getting, and how did you arrive at it ? $\endgroup$
    – true blue anil
    May 28 at 7:25
  • $\begingroup$ I only got 3C1*7C3 + 3C2 * 7C2 + 3C3 * 7C1 = 175 $\endgroup$
    – yuyu
    May 28 at 7:27
  • $\begingroup$ Once you choose one of the three eligible children as team captain, the other nine can still fill the other three team positions, with order being unimportant $ \ (252 \ = \ 3 · 84) \ \ . $ $\endgroup$
    – boojum
    May 28 at 8:06

2 Answers 2

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Hint

You have forgotten that the $3$ who can become captains can also play chess. If you correct for that, you should get the right answer.

The simplest way is

[Choose captain]*[Choose $3$ more from all 9 remaining]

$\dbinom31\dbinom93 = 252$

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  • $\begingroup$ I already included them inside the group of 4? $\endgroup$
    – yuyu
    May 28 at 7:53
  • $\begingroup$ Since a long answer has been posted, I'm posting a short method $\endgroup$
    – true blue anil
    May 28 at 9:03
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Case 1: Choosing 1 captain from 3 eligible captains and 3 players from 7 children (who aren't eligible to be captain).

Number of teams $$ = \binom{3}{1} * \binom{7}{3} $$

Case 2: Choosing 2 captains from 3 eligible captains and 2 players from 7 students. For every selection of children we have to choose a captain of the team, which means that every selection can form two possible with different captains.

Number of teams $$ = 2 * \binom{3}{2} * \binom{7}{2}$$

Case 3: Choosing 3 captains from 3 eligible captains and 1 player from 7 children. This time we can choose the captain from 3 eligible children.

Number of teams $$ = 3 * \binom{3}{3} * \binom{7}{1}$$

Therefore the total number of ways of forming a team

$$ = \binom{3}{1} * \binom{7}{3} + 2 * \binom{3}{2} * \binom{7}{2} + 3 * \binom{3}{3} * \binom{7}{1} = 252 $$

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