Question

A guard of 12 men is formed from a group of n soldiers in all possible ways. Find:
[i] The number of times two particular soldiers A and B are together on guard.
[ii] The number of times three particular soldiers C, D and E are together on guard.
[iii] Also find n if it is found that the A and B are three times as often together on guard as C, D and E are.

Answer 1

Hint: When A and B are together on guard, we have to select ten more soldiers from remaining n-2 soldiers. When C, D and E are together on guard, we have to select nine more soldiers from remaining n-3 soldiers. The number of ways in which r people are selected from a group of n people $={}^{n}{{C}_{r}}$.

Complete Complete step by step answer:
The number of ways in which A and B remain in guard together = number of ways in which we can select ten soldiers among n-2 soldiers.
This is because since A and B have to be together and also on guard, we need to select both of them and to complete selection we have to select ten more soldiers out of the remaining n- 2 soldiers.
Hence the number of ways in which A and B remain in guard together $={}^{n-2}{{C}_{10}}$
The number of ways in which C, D and E remain in guard together = number of ways in which we can select nine soldiers among n-3 soldiers.
This is because, since C, D and E have to be together and also on guard we need to select all three of them and to complete selection we have to select nine more soldiers out of the remaining n- 3 soldiers.
Hence, the number of ways in which C, D and E remain in guard together $={}^{n-3}{{C}_{9}}$
Given that
The number of ways in which A and B remain in guard together is twice the number of ways in which C, D and E remain in guard together.
Hence, we have
${}^{n-2}{{C}_{10}}=3\times {}^{n-3}{{C}_{9}}$
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Using, we get
\[\begin{align}
  & \dfrac{\left( n-2 \right)!}{10!\left( n-2-10 \right)!}=3\times \dfrac{\left( n-3 \right)!}{9!\left( n-3-9 \right)!} \\
 & \Rightarrow \dfrac{\left( n-2 \right)!}{10!\left( n-12 \right)!}=3\times \dfrac{\left( n-3 \right)!}{9!\left( n-12 \right)!} \\
 & \Rightarrow \dfrac{\left( n-2 \right)!}{\left( n-3 \right)!}=3\times \dfrac{10!}{9!} \\
 & \Rightarrow \dfrac{\left( n-2 \right)\left( n-3 \right)!}{\left( n-3 \right)!}=3\times \dfrac{10!}{9!} \\
 & \Rightarrow n-2=30 \\
 & \Rightarrow n=32 \\
\end{align}\]

Note: When making a selection of items if we select A and B, it can be done in 1 way. However, if we were arranging things then after selection, we need to permute also, and hence if we select A and B, it can be done in $1\times 2!$ ways. It has to be noted that the formation of a group is a selection and not a permutation since the order in which the soldiers are present in the group is immaterial. This is why the selection of A and B or C, D and E did not show up in the calculation of the number of ways in which group was formed.