Five people of different heights will be photographed
standing in a single row. Given that the tallest person
must stand in the middle, how many different
arrangements of the 5 people are possible?
F.
13
G.
18
H.
24
J.
120
K.
256
We can split the problem into two cases: when the tallest person is an even number of positions away from the end, and when the tallest person is an odd number of positions away from the end.
Case 1: The tallest person is 2 positions away from the end:
In this case, there are 2 possible positions for the tallest person (either in the second or fourth position). Once the tallest person is placed, there are 2 remaining positions for the shortest person. The other three people can then be arranged in $3! = 6$ ways. So, in total, there are $2 \cdot 2 \cdot 6 = 24$ arrangements in this case.
Case 2: The tallest person is 4 positions away from the end:
In this case, there is only 1 position for the tallest person (in the third position). Once the tallest person is placed, there is 1 remaining position for the shortest person. The other three people can then be arranged in $3! = 6$ ways. So, in total, there are $1 \cdot 1 \cdot 6 = 6$ arrangements in this case.
Adding the two cases together, we get $24 + 6 = \boxed{\textbf{(J) } 30}$ different arrangements possible.