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

- $C(n,r)=\large\frac{n!}{r!(n-r)!}$

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.

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