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The value of \( {}^{7}P_{3} \) is 210.
Solución
1. The permutation formula is given by:
\[
{}^{n}P_{r} = \frac{n!}{(n-r)!}.
\]
2. Substitute \( n = 7 \) and \( r = 3 \) into the formula:
\[
{}^{7}P_{3} = \frac{7!}{(7-3)!} = \frac{7!}{4!}.
\]
3. Expand the factorials:
\[
7! = 7 \times 6 \times 5 \times 4!
\]
Thus,
\[
{}^{7}P_{3} = \frac{7 \times 6 \times 5 \times 4!}{4!} = 7 \times 6 \times 5.
\]
4. Multiply to get the final result:
\[
7 \times 6 \times 5 = 210.
\]
The value of \( {}^{7}P_{3} \) is \( 210 \).
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Simplifique esta solución Extra Insights
To find the value of \( { }^{7} P_{3} \), we use the formula for permutations, which is given by: \[ ^{n} P_{r} = \frac{n!}{(n - r)!} \] In this case, \( n = 7 \) and \( r = 3 \). Plugging in the values, we have: \[ ^{7} P_{3} = \frac{7!}{(7 - 3)!} = \frac{7!}{4!} \] Calculating \( 7! = 7 \times 6 \times 5 \times 4! \), we can cancel \( 4! \): \[ ^{7} P_{3} = \frac{7 \times 6 \times 5 \times 4!}{4!} = 7 \times 6 \times 5 \] Now, multiplying those together: \[ 7 \times 6 = 42 \] \[ 42 \times 5 = 210 \] Therefore, \( { }^{7} P_{3} = 210 \).
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