2. Seating 8 people in a row at a football game: 5 adults, 3 children

Key takeaway: This is a counting problem about arranging 8 distinct people in a line. The key idea is to count all possible linear arrangements of 5 adults and 3 children.

Step 1: Count the people

There are 8 people total:

• 5 adults
• 3 children

Step 2: Arrange 8 distinct people in a row

If all 8 people are different, the number of linear arrangements is

$8!$

Step 3: Compute the value

$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$

Final answer

$\boxed{40320}$

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Pitfalls the pros know 👇 A common mistake is to use combinations instead of arrangements. Because the people are sitting in a row, order matters, so this is a permutation problem, not just a selection problem.

What if the problem changes? If some people were identical within a group, the count would change. For example, if the 5 adults were not distinct from one another and the 3 children were not distinct from one another, you would divide by repeated-factor counts. But with 8 different people, the answer is simply $8!$.

Tags: permutation, factorial, linear arrangement