Seven runners are competing in a race where 3 of them will earn medals for finishing first, second, and third. How many unique ways are there to arrange 3 of the 7 runners in first, second, and third place?

Permutations of 7 runners: We are looking to find the number of permutations of 7 runners taken 3 at a time, since the order in which they finish is important (first, second, and third are distinct positions). The formula for permutations of n items taken r at a time is P(n, r) = n! / (n - r)!. Here, n = 7 (total runners) and r = 3 (positions to fill).Calculating the factorial of n: First, we calculate the factorial of n, which is 7! (7 factorial). 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1.Calculating the factorial of (n - r): Next, we calculate the factorial of (n - r), which is (7 - 3)! or 4! (4 factorial). 4! = 4 × 3 × 2 × 1.Using the permutation formula: Now, we use the permutation formula to find the number of unique ways to arrange 3 of the 7 runners. P(7, 3) = 7! / (7 - 3)! = 7! / 4!.Simplifying the expression: We simplify the expression by canceling out the common terms in the numerator and the denominator. P(7, 3) = (7 × 6 × 5 × 4 × 3 × 2 × 1) / (4 × 3 × 2 × 1) = 7 × 6 × 5.Calculating the product: Finally, we calculate the product of the remaining terms. P(7, 3) = 7 × 6 × 5 = 42 × 5 = 210. So, there are 210 unique ways to arrange 3 of the 7 runners in first, second, and third place.

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Seven runners are competing in a race where 3 of them will earn medals for finishing first, second, and third.
How many unique ways are there to arrange 3 of the 7 runners in first, second, and third place?

Seven runners are competing in a race where $3$ of them will earn medals for finishing first, second, and third. $\newline$ How many unique ways are there to arrange $3$ of the $7$ runners in first, second, and third place?

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Q. Seven runners are competing in a race where

3

of them will earn medals for finishing first, second, and third.

\newline

How many unique ways are there to arrange

3

of the

7

runners in first, second, and third place?

Permutations of $7$ runners: We are looking to find the number of permutations of $7$ runners taken $3$ at a time, since the order in which they finish is important (first, second, and third are distinct positions). $\newline$ The formula for permutations of $n$ items taken $r$ at a time is $P(n, r) = \frac{n!}{(n - r)!}$ . $\newline$ Here, $n = 7$ (total runners) and $r = 3$ (positions to fill).
Calculating the factorial of $n$ : First, we calculate the factorial of $n$ , which is $7!$ ( $7$ factorial). $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ .
Calculating the factorial of $(n - r)$ : Next, we calculate the factorial of $(n - r)$ , which is $(7 - 3)!$ or $4!$ ( $4$ factorial). $\newline$ $4! = 4 \times 3 \times 2 \times 1.$
Using the permutation formula: Now, we use the permutation formula to find the number of unique ways to arrange $3$ of the $7$ runners. $\newline$ $P(7, 3) = \frac{7!}{(7 - 3)!} = \frac{7!}{4!}.$
Simplifying the expression: We simplify the expression by canceling out the common terms in the numerator and the denominator. $\newline$ $P(7, 3) = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} = 7 \times 6 \times 5$ .
Calculating the product: Finally, we calculate the product of the remaining terms. $\newline$ $P(7, 3) = 7 \times 6 \times 5 = 42 \times 5 = 210$ . $\newline$ So, there are $210$ unique ways to arrange $3$ of the $7$ runners in first, second, and third place.