We wish to select 6 persons from 8,


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?


We know that,


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

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,



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