Question

In a group of boys, the number of arrangements of 5 boys is 20 times the number of arrangements of 3 boys. The number of boys in the group is, choose the correct option.
A. 7
B. 8
C. 9
D. 10

Answer 1

Hint: For arrangement we use the permutation \[{}^{n}{{P}_{r}}\] of n distinct objects in r ways. So form an equation for arrangements of 5 boys and arrangements of 3 boys, to find the total number of boys (n)

Complete step-by-step answer:
In the question, we have to find the number of boys in the group if the number of arrangements of 5 boys is 20 times the number of arrangements of 3 boys. So, let the number of boys in the group is n. Also, here we know that for arrangement of n distinct objects in r ways we use the permutation and is written as \[{}^{n}{{P}_{r}}\] and this is further written as:
\[\Rightarrow {}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}\]
So here we will find the number of arrangements of 5 boys from the group of n boys, as follows:
Here r=5,
\[\begin{align}
  & \Rightarrow {}^{n}{{P}_{5}}=\dfrac{n!}{(n-5)!} \\
 & \Rightarrow {}^{n}{{P}_{5}}=\dfrac{n(n-1)(n-2)(n-3)(n-4)(n-5)!}{(n-5)!} \\
 & \Rightarrow {}^{n}{{P}_{5}}=n(n-1)(n-2)(n-3)(n-4)\,\,\,\,\,\,\,\,(eq-1) \\
\end{align}\]
Next, we will find the number of arrangements of 3 boys from the group of n boys, as follows:
Here, r=3
\[\begin{align}
  & \Rightarrow {}^{n}{{P}_{3}}=\dfrac{n!}{(n-3)!} \\
 & \Rightarrow {}^{n}{{P}_{3}}=\dfrac{n(n-1)(n-2)(n-3)!}{(n-3)!} \\
 & \Rightarrow {}^{n}{{P}_{3}}=n(n-1)(n-2)\,\,\,\,\,\,\,(eq-2) \\
\end{align}\]
Next, it is given that number of arrangements of 5 boys is 20 times the number of arrangements of 3 boys, so this will form an equation as follows:
\[\begin{align}
  & \Rightarrow {}^{n}{{P}_{5}}=20{}^{n}{{P}_{3}} \\
 & \Rightarrow n(n-1)(n-2)(n-3)(n-4)=20(n)(n-1)(n-2)\, \\
\end{align}\]
This is because the number of arrangements of 3 boys is 20 times less than the number of arrangements of 5 boys. Now we will solve the equation above to get the value of n, as shown below:
\[\begin{align}
  & \Rightarrow n(n-1)(n-2)(n-3)(n-4)=20(n)(n-1)(n-2) \\
 & \Rightarrow (n-3)(n-4)=20 \\
 & \Rightarrow {{n}^{2}}-7n+12=20\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because \left( a+b \right)\left( c+d \right)=ac+ad+bc+bd \\
\end{align}\]
Now, we will solve this quadratic equation using the formula \[n=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a},where\,\,a{{n}^{2}}+bn+c=0\] is the quadratic equation. So comparing the above equation with the general form, we get, \[a=1,b=-7,c=-8\]. So, the value of n will be:
\[\begin{align}
  & \Rightarrow n=\dfrac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-4\cdot \;1\left( -8 \right)}}{2\cdot \;1} \\
 & \Rightarrow n=\dfrac{7\pm \sqrt{81}}{2} \\
 & \Rightarrow n=8,n=-1 \\
\end{align}\]
So out of the given options, we have option B is 8
Hence, option B is the correct answer.

Note: We will use permutation \[{}^{n}{{P}_{r}}\]and not combination \[{}^{n}{{C}_{r}}\] in this question. The reason is that we are calculating the number of ways of arrangements and not the number of ways of selection.