Question

There are 10 balls of different colours. In how many ways is it possible to select 7 of them so as to exclude the white and the black ball?
A. 8
B. 7
C. 16
D. 20

Answer 1

Hint: We here have to select 7 balls out of 10 balls such that none of them is black or white. For this, we will first subtract from the total number of balls the number of balls which cannot be selected. Then we will find the number of ways of selecting 7 balls out of the leftover balls. For this, we will use the formula used for calculating ‘r’ objects out of ‘n’ different objects given as $^{n}{{C}_{r}}$ which is expanded as $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Hence, using this, we will get the required answer.

Complete step by step answer:
Here, we have been given 10 balls out of which we have to select 7 balls with and the condition given to us is that none of the selected balls should be black or white in colour.
Thus, out of 10, there are 2 balls which we shouldn’t be selecting. Thus, the number of balls out of which we can select 7 balls are given as:
$10-2=8$
Thus, we can select 7 balls out of 8 balls.
Now, we know that when we have to select ‘r’ objects from ‘n’ different objects, the number of ways to do that is given by the formula $^{n}{{C}_{r}}$.
Here, we can see that:
$\begin{align}
  & n=8 \\
 & r=7 \\
\end{align}$
Thus, putting these values in the above mentioned formula, we get:
$\begin{align}
  & ^{n}{{C}_{r}} \\
 & {{\Rightarrow }^{8}}{{C}_{7}} \\
\end{align}$
Now, we know that the expansion of $^{n}{{C}_{r}}$ is given as:
$^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Thus, we get our required value as:
$\begin{align}
  & ^{8}{{C}_{7}} \\
 & \Rightarrow \dfrac{8!}{7!\left( 8-7 \right)!} \\
 & \Rightarrow \dfrac{8!}{7!1!} \\
\end{align}$
Now, we know that $n!$ is given as $n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)...3.2.1$
This can also be written as:
$\begin{align}
  & n! \\
 & \Rightarrow n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)...3.2.1 \\
 & \Rightarrow n\left( \left( n-1 \right)\left( n-2 \right)\left( n-3 \right)...3.2.1 \right) \\
 & \Rightarrow n\left( \left( n-1 \right)! \right) \\
 & \Rightarrow n\left( n-1 \right)! \\
\end{align}$
Thus, we can write 8! as:
$8\times 7!$
Now, putting this in our expansion of $^{8}{{C}_{7}}$ we get:
$\begin{align}
  & \dfrac{8!}{7!1!} \\
 & \Rightarrow \dfrac{8\times 7!}{7!\times 1} \\
 & \therefore 8 \\
\end{align}$
Thus, the number of ways in which we can select 7 balls out of 10 balls such that none of those balls is the black ball or the white ball is 8.

So, the correct answer is “Option A”.

Note: We here have used $^{n}{{C}_{r}}$ to calculate the required number of ways. But remember, this is only applicable when n out of n objects are distinct. If ‘p’ objects out of the ‘n’ objects are the same, then the resultant is also divided by p!. Hence, if ‘r’ objects are to be selected out ‘n’ objects among which ‘p’ are similar, then the number of ways to do that is given as $\dfrac{^{n}{{C}_{r}}}{p!}$.