Question

In a textbook of mathematics there are three exercises A, B and C consisting of 12, 18 and 10 questions respectively. In how many ways can 3 questions be selected choosing one question from each exercise?

Answer 1

Hint: We will use the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ for choosing the questions from each of the exercises and then we will multiply the respective answer in the three cases after choosing 1 question and that will be the final answer.

Complete step-by-step answer:
Let’s start our solution,
Combination means that in how many ways we can choose from the given number of objects.Now if we have n different objects and from them we need to pick r objects, then the formula is
${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
Now for A,
We have n = 12, and r = 1
Now substituting the values in ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ we get,
$\begin{align}
  & =\dfrac{12!}{1!\left( 12-1 \right)!} \\
 & =\dfrac{12\times 11!}{11!} \\
 & =12 \\
\end{align}$
Now for B,
We have n = 18, and r = 1
Now substituting the values in ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ we get,
$\begin{align}
  & =\dfrac{18!}{1!\left( 18-1 \right)!} \\
 & =\dfrac{18\times 17!}{17!} \\
 & =18 \\
\end{align}$
Now for C,
We have n = 10, and r = 1
Now substituting the values in ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ we get,
$\begin{align}
  & =\dfrac{10!}{1!\left( 10-1 \right)!} \\
 & =\dfrac{10\times 9!}{9!} \\
 & =10 \\
\end{align}$
Now the answer will be $18\times 12\times 10=2160$

Note:We have used the formula of combination ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ and this must be remembered and kept in mind while solving the question. We can also look at it in this way that we have to select the three questions from three exercises such that we have to choose one question from one exercise. So, we can directly get that we have to multiply the number of questions in the given exercises and then we will get the same answer as above.