find the value of `""^(10)C_5`

Options

A. 525
B. 126
C. 252
D. 50

Answer

To find the value of \( \binom{10}{5} \), we will use the formula for combinations:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

In this case, \( n = 10 \) and \( r = 5 \). Therefore, we can write:

\[ \binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5! \cdot 5!} \]

Next, we can expand \( 10! \) as follows:

\[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5! \]

Now, substituting this back into our combination formula gives:

\[ \binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \cdot 5!} \]

The \( 5! \) in the numerator and denominator cancels out:

\[ \binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5!} \]

Now we need to calculate \( 5! \):

\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]

Now substituting \( 5! \) back into our equation gives:

\[ \binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{120} \]

Next, we calculate the numerator:

\[ 10 \times 9 = 90 \]
\[ 90 \times 8 = 720 \]
\[ 720 \times 7 = 5040 \]
\[ 5040 \times 6 = 30240 \]

Now substituting this back into our equation gives:

\[ \binom{10}{5} = \frac{30240}{120} \]

Now we perform the division:

\[ \frac{30240}{120} = 252 \]

Thus, the value of \( \binom{10}{5} \) is:

\[ \boxed{252} \]