To find the coefficient of the \(a^4b^2\) term in the binomial expansion of \((a+b)^6\), you can use the Binomial Theorem, which states:
\[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
Here, \(n\) is the total exponent (in this case, 6), and \(k\) is the number of times \(b\) is included in the term.
For the term \(a^4b^2\):
- The exponent of \(a\) is 4, which means \(n-k=4\); given \(n=6\), we find \(k=2\) since \(6-2=4\).
- The exponent of \(b\) is 2, which corresponds to \(k=2\).
The coefficient for the term \(a^4b^2\) can be calculated using the binomial coefficient \(\binom{n}{k} = \binom{6}{2}\):
\[ \text{Coefficient of } a^4b^2 = \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2! \cdot 4!} = \frac{6 \times 5}{2 \times 1} = 15. \]
In summary, to find the coefficient for the \(a^4b^2\) term, he would use:
\[ \binom{6}{2} \]
So, the answer is \(6C2\) (the second option in your list).