A question paper divided into two groups consisting of 3 and 4 questions respectively carries the note “It is not necessary to answer all the questions. One question must be answered from each group”. In how many ways you can select these questions?
(a) 10
(b) 11
(c) 12
(d) 13

Hint: From group 1, we can write either question number 1 or 2 or 3. There are $^{3}{{C}_{1}}$ ways of selecting questions from group 1. Similarly, from group 2, we can write either question number 1 or 2 or 3 or 4. There are $^{4}{{C}_{1}}$ ways of selecting questions from group 2. We can find the required solution by multiplying the numbers of ways of selecting questions from group 1 and the numbers of ways of selecting questions from group 2.

Complete step by step answer:
We are given that the question paper is divided into two groups consisting of 3 and 4 questions. We are also given that one question must be answered from each group. From group 1, we can write either question number 1 or 2 or 3. We can write the number of ways of selecting one question from group 1 as
${{\Rightarrow }^{3}}{{C}_{1}}$
Similarly, from group 2, we can write either question number 1 or 2 or 3 or 4. We can write the number of ways of selecting one question from group 2 as
${{\Rightarrow }^{4}}{{C}_{1}}$
Therefore, we can find the number of ways of selecting these questions by multiplying the numbers of ways of selecting questions from group 1 and the numbers of ways of selecting questions from group 2.
The number of ways of selecting the questions ${{=}^{3}}{{C}_{1}}{{\times }^{4}}{{C}_{1}}$
We know that $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ . Therefore, we can write the above combination as
The number of ways of selecting the questions $=\dfrac{3!}{\left( 3-1 \right)!1!}\times \dfrac{4!}{\left( 4-1 \right)!1!}$
$\Rightarrow $ The number of ways of selecting the questions $=\dfrac{3!}{2!1!}\times \dfrac{4!}{3!1!}$
We know that $n!=n\times \left( n-1 \right)!$ .
$\Rightarrow $ The number of ways of selecting the questions $=\dfrac{3\times 2!}{2!}\times \dfrac{4\times 3!}{3!}$
We have to cancel the common terms.
$\Rightarrow $ The number of ways of selecting the questions $=3\times 4=12$

So, the correct answer is “Option C”.

Note: Students have a chance of making mistake by adding $^{3}{{C}_{1}}$ and $^{4}{{C}_{1}}$ instead of multiplying them. Students must be thorough with the formula of combination. They may get confused with the formula for permutation and combination. The formula for permutation is $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ . We need to find the result of $^{3}{{C}_{1}}$ and $^{4}{{C}_{1}}$ using the combination formula if we know the standard result that $^{n}{{C}_{1}}=n$ .