The question says:
We wish to select $6$ people from $8$, if the person $\text{A}$ is chosen, then $\text{B}$ must be chosen. In how many ways selections can be made?
If we need to keep $\text{A}$ and $\text{B}$ in our group, we need to choose $(6 – 2) = 4$ more from $(8 – 2) = 6$, as we already have $2$ people included.
The number of ways to choose : $^6C_4 = 15$.
Now, what if we exclude both $\text{A}$ and $\text{B}$?
We have $(8 – 2) = 6$ people to choose from, and we need to choose $6$.
Number of ways to do so : $^6C_6 = 1$.
Both are mutually exclusive events, so Total number of selections : $15 + 1 = 16$.
But the answer key reports the answer is $22$.
What did I miss? Do we have to take $\text{B}$ in our team without $\text{A}$? The question says if $\text{A}$ is chosen, $\text{B}$ must be chosen, but there are no strict instructions to do the opposite one.
Actually, let's try.
We want to keep $\text{B}$, without keeping $\text{A}$. Then, we have $(8 – 2) = 6$ people to choose from, and we need to choose $(6- 1) = 5$.
The number of ways to do it : $^6C_5 = 6$.
The answer now matches, but who is wrong? Me, or the textbook answer? I found the last addition kind of irrational.
Any help would be deeply appreciated.
Edit: While Excluding $\text{A}$, $\text{B}$ has already been chosen, so we have remaining $6$ to choose from, not $7$. It was an error I made in the first version.
Best Answer
As pointed out by Mr. Shore and Mr. Taussig, The instructions of the questions allows us to perform the calculation of "Including $\text{B}$ but not $\text{A}$", so We have $16 + 6 = 22$ ways of choosing.
Thanks for clearing my doubt on this one.