Team of 3 children, & a captain is required. 10 children can play chess of whom only 3 qualified to act captain. Find number of ways to select team.

Question

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.

Answer 1

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.

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The simplest way is

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

$\dbinom31\dbinom93 = 252$

Answer 2

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

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