# Easy Permutation-Combination Solved QuestionAptitude Discussion

 Q. In an examination paper, there are two groups each containing $4$ questions. A candidate is required to attempt $5$ questions but not more than $3$ questions from any group. In how many ways can $5$ questions be selected?
 ✖ A. 24 ✔ B. 48 ✖ C. 96 ✖ D. 64

Solution:
Option(B) is correct

$5$ questions can be selected in the following ways:

$2$ question from first group and 3 question from second group

Or

$3$ question from first group and 2 question from second group.

$({^4C_2} × {^4C_3}) + ({^4C_3} × {^4C_2})$

$= 24 + 24$

$=\textbf{48}$

## (1) Comment(s)

RAKESH
()

Here, we have 2 groups each having 5 questions and condition is we should not attempt more than 3 question from each group so

By using the combination

${^4C_3} \times {^4C_2}+{^4C_2} \times {^4C_3}=48$