To find the coefficient of the sixth term in the binomial expansion of \((a+b)^9\), we can use the Binomial Theorem, which states:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
Here, \(n = 9\), and we are looking for the sixth term of the expansion. Recall that the terms in the expansion are indexed starting from \(k=0\). Therefore, the sixth term corresponds to \(k=5\) (since the first term, \(k=0\), is the 1st term).
The general term (the \(k\)-th term) in the expansion can be expressed as:
\[ T_k = \binom{n}{k} a^{n-k} b^k \]
For \(n = 9\) and \(k = 5\), we have:
\[ T_5 = \binom{9}{5} a^{9-5} b^5 = \binom{9}{5} a^4 b^5 \]
Now, we need to calculate \(\binom{9}{5}\):
\[ \binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!} \]
We can simplify \(9!\):
\[ 9! = 9 \times 8 \times 7 \times 6 \times 5! \]
Thus,
\[ \binom{9}{5} = \frac{9 \times 8 \times 7 \times 6}{4!} = \frac{9 \times 8 \times 7 \times 6}{24} \]
Calculating the numerator:
\[ 9 \times 8 = 72 \] \[ 72 \times 7 = 504 \] \[ 504 \times 6 = 3024 \]
Now, we calculate the complete binomial coefficient:
\[ \binom{9}{5} = \frac{3024}{24} = 126 \]
Finally, the coefficient of the sixth term \(T_5\) is
\[ \boxed{126} \]
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