Vocal group no two singer stand next to each other? A vocal group consisting of alf,bill,cal,deb,eve, and fay are deciding how to arrange themselves from left to right on a stage How many way to do this if
Alf and Fay are the least skilled singer and tend to driff of key so they should not stand next to each other?
Since Alf and fay cannot stand next to each other. And you have 6 place to stand
You have four places where four possible people can stand and 2 places in which 2 people can stand.
So  would it be $(4) 4!)+2(2!)$
 A: There are always many ways to do the counting. Here is one. Even though singers should not really sit, let us imagine that there are $4$ chairs for the talented singers, lined up like this:
$$ \text{T}\qquad  \text{T}\qquad  \text{T}\qquad  \text{T}$$
That leaves $5$ "gaps" for our no talent pair to drag their chairs into, $3$ between chairs, and $2$ endgaps.
We must choose $2$ of these gaps. This can be done in $\binom{5}{2}$ ways. 
For each choice, there are $2!$ ways to permute our no-talent pair, and then $4!$ ways to permute the talented ones, for a total of $\binom{5}{2}(2!)(4!)$. 
Another way: Imagine a line of $6$ people. There are $\binom{6}{2}=15$ ways to choose the positions to be occupied by a set of $2$ people (we are not yet choosing who occupies each position). Of these, $5$ are forbidden, leaving a total of $10$. Now multiply by $2!4!$ for the usual reason.
A: If we group $AF$ as the same group, then in total we would get $5!2!$ combinations. The $2!$ accounts for permuting $AF.$
Subtracting this from the possible combinations yields $$6! - 5!2! = 480.$$
