Verified Answer :

  • The expression \({}_7P_4\) represents the number of ways to choose and arrange 4 items from a set of 7.
  • Use the formula \({}_nP_r = \frac{n!}{(n-r)!}\) with \(n=7\) and \(r=4\).
  • Simplify to \(\frac{7!}{3!} = 7 \times 6 \times 5 \times 4\).
  • Compute the product to get the final result: \(\boxed{840}\).

Explanation

  1. Understand the permutation formula

The expression \({}_7P_4\) represents a permutation, which is the number of ways to arrange a specific number of items selected from a larger set where the order matters. The general formula for calculating permutations of \(n\) objects taken \(r\) at a time is given by:
\[{}_nP_r = \frac{n!}{(n-r)!}\]
In this problem, we are looking for \({}_7P_4\), which means we have a set of \(7\) items (\(n=7\)) and we want to select and arrange \(4\) of them (\(r=4\)).

  1. Substitute and expand the expression

Using the values \(n = 7\) and \(r = 4\), we substitute them into the formula:
\[{}_7P_4 = \frac{7!}{(7-4)!} = \frac{7!}{3!}\]
The factorial of a number \(n!\) is the product of all positive integers up to \(n\). Therefore, we can expand \(7!\) and \(3!\) as follows:
\[7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\]
\[3! = 3 \times 2 \times 1\]
Substituting these into our expression, we get:
\[\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1}\]

  1. Final calculation and result

Notice that the terms \(3 \times 2 \times 1\) in the numerator and the denominator cancel each other out:
\[\frac{7 \times 6 \times 5 \times 4 \times (3 \times 2 \times 1)}{(3 \times 2 \times 1)} = 7 \times 6 \times 5 \times 4\]
Now, we calculate the remaining product:
\[7 \times 6 = 42\]
\[42 \times 5 = 210\]
\[210 \times 4 = 840\]
Thus, the value of the expression is \(\boxed{840}\).

Examples

Permutations are essential in real-world scenarios where order matters, such as assigning roles in a committee or creating secure access codes. For example, if you have 7 unique finalists in a competition and need to choose 4 to receive 1st, 2nd, 3rd, and 4th place ribbons, the number of possible outcomes is calculated using permutations: \({}_7P_4 = \frac{7!}{3!} = 840\) different ways to award the prizes.