In how many ways can four people line up in a straight line if the youngest person cannot be first in line?
Consider the number of people that can fill each place in line. There are three people who could be first (the youngest is excluded). There are then three people who could be second, two people who could be third, and the last is determined. Thus, there are $3 \cdot 3 \cdot 2 = \boxed{18}$ ways in which to form a line.
18
Level 2
Counting & Probability
Problem: In how many ways can four people line up in a straight line if the youngest person cannot be first in line?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the number of ways four people can line up with the restriction that the youngest person cannot be first, we start by calculating the total arrangements without restrictions. The total number of arrangements for four people is given by the factorial of the number of people, which is 4! = 24.

Next, we need to account for the restriction. If the youngest person were to be first in line, the remaining three people can be arranged in the remaining three positions. The number of arrangements in this case is 3! = 6.

To find the number of valid arrangements where the youngest person is not first, we subtract the number of arrangements where the youngest is first from the total arrangements. Thus, we calculate:

Total arrangements without restriction - Arrangements with the youngest first = 24 - 6 = 18.

This leads us to conclude that the number of ways four people can line up in a straight line, ensuring the youngest person is not in the first position, is 18.<|END_OF_LATENT|> The final answer is $18$.