The problem asks for the number of ways 5 people can be seated in a row such that two specific people, A and B, are not seated next to each other. We can solve this using the principle of complementary counting in permutations.
First, find the total number of ways to arrange 5 distinct people in a row without any restrictions. This is calculated using the factorial of the number of people.
Total arrangements = $5!$
$ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $
There are 120 distinct ways to seat 5 people in a row.
Next, we calculate the number of arrangements where A and B *are* seated next to each other. We can treat A and B as a single combined unit or block.
Consider the block (AB). Now we have 4 items to arrange: the block (AB) and the other 3 people (C, D, E). The number of ways to arrange these 4 items is $4!$.
$ 4! = 4 \times 3 \times 2 \times 1 = 24 $
Within the block (AB), A and B can switch positions (AB or BA). There are $2!$ ways to arrange A and B within the block.
$ 2! = 2 \times 1 = 2 $
The total number of arrangements where A and B sit together is the product of these two calculations:
Arrangements (A and B together) = (Ways to arrange 4 items) $\times$ (Ways to arrange A and B within the block)
$ \text{Ways (Together)} = 4! \times 2! = 24 \times 2 = 48 $
To find the number of ways A and B are *not* together, we subtract the number of arrangements where they *are* together from the total number of arrangements.
Arrangements (A and B not together) = Total Arrangements - Arrangements (A and B together)
$ \text{Ways (Not Together)} = 120 - 48 = 72 $
The total number of ways to seat 5 people is 120. The calculation shows that there are 48 ways for A and B to sit together, and 72 ways for A and B *not* to sit together.
The value 48 corresponds to the specific scenario where A and B are adjacent in the seating arrangement.
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