# We wish to select 6 persons from 8,but if the person A is chosen,then B must be chosen.In how many ways can selection be made?

$\begin{array}{1 1}(A)\;20\\(B)\;22\\(C)\;24\\(D)\;26\end{array}$

Toolbox:
• $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Case I :A is included and B is not
To select 6 persons =$6C_6$=1
Case II : B is selected A is not selected
$\Rightarrow$Number of ways =$1\times 6C_5$
$\Rightarrow \large\frac{6!}{5!1!}$
$\Rightarrow \large\frac{6\times 5!}{5!}$$=6$
Case III:A and B both are selected
Number of ways =$1\times 1\times 6C_4$
$\Rightarrow 1\times 1\times \large\frac{6!}{4!2!}$
$\Rightarrow \large\frac{6\times 5\times 4!}{4!\times 2}$
$\Rightarrow 15$
Total number of ways =$15+6+1=22$
Hence (B) is the correct answer.