Question

Seven letters, P, Q, R, S, T, U and V, are arranged in a row. If it is required that P and Q are to be separated, how many way(s) of arrangement are there?

Answer 1

Since when these 2 near each other ( 1 is outermost ) we can simply into the problem for P and Q arrange in a row, and since there are 2 outermost ones so we need to multiply by 2. So when P or Q is at one of any outermost ones, there are : 2! * 2 = 4 ( ways ).

Here, all arangement must count once. So after that, we don't worrry about the second outermost ones because we reduce to the simpler problem. And we do this until there is 1 left ( just work for the odd ones ).

So if n is the number of objects, the formula for this is \(n!-4\left[\frac{\left(n-1\right)}{2}\right]\).

Apply n = 7; we have \(n!-4\left[\frac{\left(n-1\right)}{2}\right]=7!-4\left[\frac{\left(7-1\right)}{2}\right]=5028\)

Ans: 5028 ways