Coefficient of `x^(5)` in `(1 + x)^(10)` is `"^(10)C_(5)`.

Answer

To find the coefficient of \( x^5 \) in the expansion of \( (1 + x)^{10} \), we can apply the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

In our case, we have \( a = 1 \), \( b = x \), and \( n = 10 \). Therefore, we can write:

\[ (1 + x)^{10} = \sum_{k=0}^{10} \binom{10}{k} 1^{10-k} x^k \]

Since \( 1^{10-k} = 1 \) for any value of \( k \), the expression simplifies to:

\[ (1 + x)^{10} = \sum_{k=0}^{10} \binom{10}{k} x^k \]

We are interested in the coefficient of \( x^5 \). According to the binomial expansion, the coefficient of \( x^k \) is given by \( \binom{n}{k} \). Therefore, the coefficient of \( x^5 \) in this expansion is:

\[ \binom{10}{5} \]

Thus, the coefficient of \( x^5 \) in \( (1 + x)^{10} \) is \( \binom{10}{5} \).

### Final Answer:
The coefficient of \( x^5 \) in \( (1 + x)^{10} \) is \( \binom{10}{5} \).

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