Question

A village has 10 players. A team of 6 players is to be formed. 5 members are chosen first out of these 10 players and then the captain from the remaining players. Then the total number of ways of choosing such teams is \[\]
A.1260 \[\]
B.210 \[\]
C. $ \left( {}^{10}{{C}_{6}} \right)5! $ \[\]
D. $ {}^{10}{{C}_{4}}6 $ \[\]

Answer 1

Hint: We use the formula for selecting $ r $ objects from $ n $ distinct objects as $ {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} $ . We find the number of ways to fill the spot of 5 players by selecting 5 players from 10 and the number of ways to fill the rest 1 empty spot for captain by selecting 1 player out of the rest 5 players. We use the rule of product and multiply 2 results to get the answer. \[\]

Complete step by step answer:
We are given in the question that the village has 10 players and a team of 6 players is to be formed subjected to condition that 5 members are chosen first out of these 10 players and then the captain from the remaining players. \[\]
We first select 5 members for the team out of 10 members in $ {}^{10}{{C}_{5}} $ ways and fill the 5 spots for the team. Now there is only $ 6-5=1 $ spot left for the team and is reserved for the captain and there are is $ 10-5=5 $ players remaining to choose. We cans select the captain out of rest 5 players in $ {}^{5}{{C}_{1}} $ ways. \[\]
We know from rule of product that if we can do one thing in $ m $ ways and another thing in $ n $ ways then we can do both things simultaneously in $ mn $ ways. Since we are selecting the 5 players and the captain at once by rule of product the number of ways the team can be formed is,
\[\begin{align}
  & {}^{10}{{C}_{3}}\times {}^{5}{{C}_{1}}=\dfrac{10!}{5!\left( 10-5 \right)!}\times \dfrac{5!}{1!\left( 5-1 \right)!} \\
 & \Rightarrow {}^{10}{{C}_{3}}\times {}^{5}{{C}_{1}}=\dfrac{10!}{5!5!}\times \dfrac{5!}{1!4!} \\
 & \Rightarrow {}^{10}{{C}_{3}}\times {}^{5}{{C}_{1}}=252\times 5 \\
 & \Rightarrow {}^{10}{{C}_{3}}\times {}^{5}{{C}_{1}}=1260 \\
\end{align}\]
 So the correct option is C. \[\]

Note:
 We should be careful that the objects we are selecting must be distinct and in this case team members are distinct. We are also not replacing back the members in which case the number of selection will be $ {}^{n+r-1}{{C}_{r}} $ . We must be careful of confusion between the rule of product and rule of the sum which states that we can do wither of that if we can do one thing in $ m $ ways and another thing in $ n $ ways then we can EITHER of the things in $ m+n $ ways.