Let's consider the expansion of [tex]\((x + y)^6\)[/tex]. According to the Binomial Theorem, the general term in the expansion of [tex]\((x + y)^n\)[/tex] is given by:
[tex]\[
\binom{n}{k} x^{n-k} y^k
\][/tex]
where [tex]\(\binom{n}{k}\)[/tex] is the binomial coefficient, defined as:
[tex]\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\][/tex]
In this case, [tex]\(n = 6\)[/tex] and we are expanding [tex]\((x + y)^6\)[/tex]. We need to find the coefficients for each term in the expansion, for values of [tex]\(k\)[/tex] ranging from 0 to 6.
Thus, we need to calculate [tex]\(\binom{6}{k}\)[/tex] for [tex]\(k = 0, 1, 2, 3, 4, 5, 6\)[/tex].
Given the results are:
[tex]\[
\begin{aligned}
\binom{6}{0} &= 1, \\
\binom{6}{1} &= 6, \\
\binom{6}{2} &= 15, \\
\binom{6}{3} &= 20, \\
\binom{6}{4} &= 15, \\
\binom{6}{5} &= 6, \\
\binom{6}{6} &= 1.
\end{aligned}
\][/tex]
Therefore, the coefficients in the binomial expansion of [tex]\((x + y)^6\)[/tex] are:
[tex]\[1, 6, 15, 20, 15, 6, 1\][/tex]
Based on this breakdown, the correct option among the provided choices is:
C. [tex]\(1, 6, 15, 20, 15, 6, 1\)[/tex]
