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A team of 8 players is to be chosen from a group of 12 players. Out of the eight players one is to be elected as captain and another vice-captain. In how many ways can this be done?
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The correct answer is:
27720
Total no. of players $=12$ No. of chosen players $=8$ Number of ways to choose 8 players from 12 players
$={ }^{12} C_{8}=\frac{12 !}{8 ! 4 !}=\frac{12 \times 11 \times 10 \times 9 \times 8 !}{8 ! 4 !}=495$
Since, out of the 8 players 1 is to be elected as captain and another vice-captain Therefore number of ways to choose a captain and a vice-captain
$={ }^{8} C_{1} \times{ }^{7} C_{1}=8 \times 7=56 \quad\left(\because{ }^{n} C_{1}=n\right)$
Hence, required number of ways $=495 \times 56=27720$
$={ }^{12} C_{8}=\frac{12 !}{8 ! 4 !}=\frac{12 \times 11 \times 10 \times 9 \times 8 !}{8 ! 4 !}=495$
Since, out of the 8 players 1 is to be elected as captain and another vice-captain Therefore number of ways to choose a captain and a vice-captain
$={ }^{8} C_{1} \times{ }^{7} C_{1}=8 \times 7=56 \quad\left(\because{ }^{n} C_{1}=n\right)$
Hence, required number of ways $=495 \times 56=27720$
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