Question:
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 selections be made?
Solution:
We know that,
nCr
$=\frac{n !}{r !(n-r) !}$
According to the question,
Case 1:
If both A and B are selected =1x1x6C4
$=\frac{6 !}{4 !(6-4) !}=\frac{6 !}{4 ! 2 !}=15$
Case 2:
If neither A nor B are selected = 6C6 = 1
If B is selected but A is not selected = 1x6C5
$=\frac{6 !}{5 !(6-4) !}=6$
Adding the results of both A and B being selected, neither A nor B being selected and B being selected but A not being selected,
We get,
15+1+6=22