Question
Determine if the situation describes a permutation or a combination. Justify your reasoning. Then solve the problem.
How many ways could you award 1st, 2nd, and 3rd place medals to 30 contestants in the science fair?
How many ways could you award 1st, 2nd, and 3rd place medals to 30 contestants in the science fair?
Answers
GPT-4o mini
The situation describes a **permutation**. This is because the order in which the medals are awarded matters: being awarded 1st place is different from being awarded 2nd or 3rd place. Thus, we are concerned with the arrangement of the medals among the contestants.
To solve the problem of awarding 1st, 2nd, and 3rd place medals to 30 contestants, we can use the permutation formula for selecting and arranging \( r \) items from \( n \) total items, given by:
\[
P(n, r) = \frac{n!}{(n - r)!}
\]
In this case, \( n = 30 \) (the number of contestants) and \( r = 3 \) (the number of positions for medals).
So we calculate:
\[
P(30, 3) = \frac{30!}{(30 - 3)!} = \frac{30!}{27!} = 30 \times 29 \times 28
\]
Now, we compute that:
\[
30 \times 29 = 870
\]
\[
870 \times 28 = 24360
\]
Thus, the number of ways to award 1st, 2nd, and 3rd place medals to 30 contestants is **24,360**.
To solve the problem of awarding 1st, 2nd, and 3rd place medals to 30 contestants, we can use the permutation formula for selecting and arranging \( r \) items from \( n \) total items, given by:
\[
P(n, r) = \frac{n!}{(n - r)!}
\]
In this case, \( n = 30 \) (the number of contestants) and \( r = 3 \) (the number of positions for medals).
So we calculate:
\[
P(30, 3) = \frac{30!}{(30 - 3)!} = \frac{30!}{27!} = 30 \times 29 \times 28
\]
Now, we compute that:
\[
30 \times 29 = 870
\]
\[
870 \times 28 = 24360
\]
Thus, the number of ways to award 1st, 2nd, and 3rd place medals to 30 contestants is **24,360**.