Question

In a group of boys, the number of arrangements of 4 boys is 12 times the number of arrangements of 2 boys. Find the number of boys in the group.
A. 10
B. 8
C. 6
D. none of these

Answer 1

We assume the number of boys in the group. We try to find the process of arrangements of 4 boys out of n boys. From the given conditions of arrangements of 4 boys is 12 times the number of arrangements of 2 boys we form a quadratic equation. We solve it to get the value of n.

Complete step-by-step solution
Let’s assume that there are “n” boys in the group.
We are arranging r boys out of n boys. Here r has two values 2 and 4. $n\in \mathbb{N}$.
Now the process of arrangement involves first choosing the boys and then making the arrangement of only those boys.
The number of ways r boys can be arranged out of n boys is ${}^{n}{{P}_{r}}={}^{n}{{C}_{r}}\times n!$.
We know the formula of permutation tells us ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$.
So, the number of arrangements of 4 boys out of n boys is ${}^{n}{{P}_{4}}$.
Value of the arrangement is ${}^{n}{{P}_{4}}=\dfrac{n!}{\left( n-4 \right)!}$.
Again, the number of arrangements of 2 boys out of n boys is ${}^{n}{{P}_{2}}$.
Value of the arrangement is ${}^{n}{{P}_{2}}=\dfrac{n!}{\left( n-2 \right)!}$.
It’s given that the number of arrangements of 4 boys is 12 times the number of arrangements of 2 boys. We put that condition in the form of mathematical expression.
$\begin{aligned} & {}^{n}{{P}_{4}}=12\times {}^{n}{{P}_{2}} \\\ & \Rightarrow \dfrac{n!}{\left( n-4 \right)!}=12\times \dfrac{n!}{\left( n-2 \right)!} \\\ \end{aligned}$
We solve the equation by taking all the variables in one side.
$\begin{aligned} & \dfrac{n!}{\left( n-4 \right)!}=12\times \dfrac{n!}{\left( n-2 \right)!} \\\ & \Rightarrow \dfrac{n!\times \left( n-2 \right)!}{n!\times \left( n-4 \right)!}=12 \\\ & \Rightarrow \left( n-2 \right)\left( n-3 \right)=12 \\\ \end{aligned}$
We got a quadratic equation. We solve it and get the possible values of n.
$\begin{aligned} & \left( n-2 \right)\left( n-3 \right)=12 \\\ & \Rightarrow {{n}^{2}}-5n+6-12=0 \\\ & \Rightarrow {{n}^{2}}-5n-6=0 \\\ & \Rightarrow {{n}^{2}}-6n+n-6=0 \\\ & \Rightarrow \left( n-6 \right)\left( n+1 \right)=0 \\\ \end{aligned}$
So, the possible values are $n=-1,6$.
As the values of n can’t be negative, the value of n will be $n=6$.
So, the number of boys in the group is 6. The correct option is C.

Note: We don’t need to solve the equation ${}^{n}{{P}_{4}}=\dfrac{n!}{\left( n-4 \right)!}$ and ${}^{n}{{P}_{2}}=\dfrac{n!}{\left( n-2 \right)!}$ from the very start. We need to solve the equation where the majority of the terms get canceled out. Also, we need to always double-check the possible outcomes of n and its validity.