To determine the coefficient of the third term in the expansion of \((a + b)^6\), we will refer to Pascal's Triangle.
In general, the Binomial Theorem states that: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \] where \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\).
Here, \(n = 6\). The third term corresponds to \(k = 2\) (because the term indexing starts from \(k = 0\)). Therefore, we need to compute: \[ \binom{6}{2} a^{6-2} b^2 = \binom{6}{2} a^{4} b^{2} \]
Next, we calculate the binomial coefficient \(\binom{6}{2}\): \[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]
Thus, the third term in the expansion of \((a + b)^6\) is: \[ 15 a^{4} b^{2} \]
Consequently, the coefficient of the third term is: \[ \boxed{15} \]