Question

The number of ways 10 boys can be divided into 2 groups of 5, such that the two tallest boys are in the different groups.
(A) 70
(B) 35
(C) 252
(D) 126

Answer 1

Hint: We solve this problem by first selecting the two tallest students into two groups and counting the number of ways it can be done. Then we find the number of ways of dividing the remaining 8 students into two groups of 4 students each using the formula for the number of ways of dividing a set of n objects into m groups of m objects each, $\dfrac{n!}{\left( r! \right)\times \left( n-r \right)!\times m!}$. Then we multiply this value with 2 as above we have divided the two tallest students into two groups.

Complete step by step answer:
We are given that there are 10 boys.
We need to divide them into two groups of 5 each such that the two tallest boys are in the different groups.
So, first, let us select the two tallest students and place them into two groups. They can be divided into two groups in 2 ways.
 Then the remaining students that needed to be divided into groups is 8.
Now we need to divide the remaining students into 2 groups of 4 students each.
Now, let us consider the formula for the number of ways of dividing a set of n objects into m groups of m objects each is
$\dfrac{n!}{\left( r! \right)\times \left( n-r \right)!\times m!}$
Using this formula, we can select 4 students from the remaining 8 students in
$\begin{align}
  & \Rightarrow \dfrac{8!}{\left( 4! \right)\times \left( 8-4 \right)!\times 2!} \\
 & \Rightarrow \dfrac{8!}{\left( 4! \right)\times \left( 4! \right)\times 2!} \\
 & \Rightarrow \dfrac{5\times 6\times 7\times 8}{24\times 2} \\
 & \Rightarrow 35 \\
\end{align}$
So, the total number of ways is,
$\begin{align}
  & \Rightarrow 2\times 35 \\
 & \Rightarrow 70 \\
\end{align}$
So, the total number of ways of dividing them into two groups is 70.
Hence, the answer is Option A.

Note:
 We can also solve this question in an alternate process also.
The number of ways 10 boys can be divided into 2 groups of 5, such that the two tallest boys are in the different groups is equal to,
$\Rightarrow \left( \begin{align}
  & \text{Total}\ \text{number}\ \text{of}\ \text{ways of}\ \text{dividing } \\
 & \text{10 students}\ \text{into}\ \text{two}\ \text{groups} \\
\end{align} \right)$ $-\left( \begin{align}
  & \text{Number}\ \text{of}\ \text{ways}\ \text{in}\ \text{which}\ \text{two}\ \\
 & \text{tallest}\ \text{students}\ \text{are}\ \text{in}\ \text{same}\ \text{group} \\
\end{align} \right)$
$\Rightarrow $(Total number of ways of dividing 10 students into 2 groups)
– (Number of ways in which two tallest students are in same group)
Now, let us consider the formula for number of ways of dividing a set of n objects into m groups of m objects each are
$=\dfrac{n!}{\left( r! \right)\times \left( n-r \right)!\times m!}$
Using this formula, we get,
$\Rightarrow \left( \begin{align}
  & \text{Total}\ \text{number}\ \text{of}\ \text{ways of}\ \text{dividing } \\
 & \text{10 students}\ \text{into}\ \text{two}\ \text{groups} \\
\end{align} \right)$ $=\dfrac{10!}{5!\times 5!\times 2!}$
$\Rightarrow \left( \begin{align}
  & \text{Total}\ \text{number}\ \text{of}\ \text{ways of}\ \text{dividing } \\
 & \text{10 students}\ \text{into}\ \text{two}\ \text{groups} \\
\end{align} \right)$ $=\dfrac{6\times 7\times 8\times 9\times 10}{120\times 2}$
$\Rightarrow \left( \begin{align}
  & \text{Total}\ \text{number}\ \text{of}\ \text{ways of}\ \text{dividing } \\
 & \text{10 students}\ \text{into}\ \text{two}\ \text{groups} \\
\end{align} \right)$ $=126$
Now let us find the number of ways of dividing 10 students into 2 groups in which two tallest students are in the same group.
Now, let us consider the formula for number of ways of selecting r objects from a set of n objects are
${}^{n}{{C}_{r}}=\dfrac{n!}{\left( r! \right)\times \left( n-r \right)!}$
Then number of ways is equal to number of ways of selecting 3 students from 8 remaining students.
$\begin{align}
  & \Rightarrow {}^{8}{{C}_{3}}=\dfrac{8!}{\left( 3! \right)\times \left( 8-3 \right)!} \\
 & \Rightarrow {}^{8}{{C}_{3}}=\dfrac{8!}{\left( 3! \right)\times \left( 5! \right)} \\
 & \Rightarrow {}^{8}{{C}_{3}}=\dfrac{6\times 7\times 8}{6}=56 \\
\end{align}$
So, the number of ways in which they can be divided into two groups in which the tallest students are in different groups is equal to,
$\Rightarrow \left( \begin{align}
  & \text{Total}\ \text{number}\ \text{of}\ \text{ways of}\ \text{dividing } \\
 & \text{10 students}\ \text{into}\ \text{two}\ \text{groups} \\
\end{align} \right)$$-\left( \begin{align}
  & \text{Number}\ \text{of}\ \text{ways}\ \text{in}\ \text{which}\ \text{two}\ \\
 & \text{tallest}\ \text{students}\ \text{are}\ \text{in}\ \text{same}\ \text{group} \\
\end{align} \right)$
$\begin{align}
  & \Rightarrow 126-56 \\
 & \Rightarrow 70 \\
\end{align}$
Hence required number of ways is 70.
Hence answer is Option A.