Question

your answer is incorrect.

• $_{10}\mathrm{c}_3$: one of the digits in your answer is wrong. check your work carefully

evaluate $_{4}\mathrm{p}_3$ and $_{10}\mathrm{c}_3$.
your answer
$_{4}\mathrm{p}_3 = 24$
$_{10}\mathrm{c}_3 = \square$

Explanation:

Step1: Recall permutation formula

The formula for permutation \( _nP_r=\frac{n!}{(n - r)!} \). For \( _4P_3 \), we have \( n = 4 \), \( r=3 \). So \( _4P_3=\frac{4!}{(4 - 3)!}=\frac{4!}{1!}=\frac{4\times3\times2\times1}{1}=24 \) (this part was correct).

Step2: Recall combination formula

The formula for combination \( _nC_r=\frac{n!}{r!(n - r)!} \). For \( _{10}C_3 \), we have \( n = 10 \), \( r = 3 \). So \( _{10}C_3=\frac{10!}{3!(10 - 3)!}=\frac{10!}{3!7!} \).

Step3: Simplify the combination formula

We know that \( n!=n\times(n - 1)\times\cdots\times1 \), so \( \frac{10!}{3!7!}=\frac{10\times9\times8\times7!}{3\times2\times1\times7!} \). The \( 7! \) terms cancel out. Then we calculate \( \frac{10\times9\times8}{3\times2\times1}=\frac{720}{6}=120 \).

Answer:

\( _4P_3 = 24 \), \( _{10}C_3=120 \)