To find the coefficient of the sixth term in the binomial expansion of \((a + b)^9\) using the Binomial Theorem, we start by recalling the general formula for the \(k\)-th term in the expansion of \((a + b)^n\):
\[ T_k = \binom{n}{k-1} a^{n-(k-1)} b^{k-1} \]
where \(n\) is the exponent, \(k\) is the term number, and \(\binom{n}{k-1}\) is the binomial coefficient.
In this case, \(n = 9\) and we want the sixth term, which means \(k = 6\).
Using the formula:
\[ T_6 = \binom{9}{6-1} a^{9-(6-1)} b^{6-1} \]
Calculating the components:
- \( \binom{9}{6-1} = \binom{9}{5} \)
- The exponent of \(a\) is \(9 - (6 - 1) = 9 - 5 = 4\).
- The exponent of \(b\) is \(6 - 1 = 5\).
Now we compute \(\binom{9}{5}\):
\[ \binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9!}{5! \cdot 4!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 126 \]
Putting it all together, the sixth term is:
\[ T_6 = \binom{9}{5} a^4 b^5 = 126 a^4 b^5 \]
Thus, the coefficient of the sixth term in the binomial expansion of \((a + b)^9\) is:
\[ \boxed{126} \]
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