Given a problem such as "find the coefficient of $a^2b^6$ for $(a+3b+2)^{10}$," how would I go about doing this?
I know the multinomial theorem, but I'm not sure how to approach this problem given that $b$ has a coefficient, as well as the fact that the powers of $a$ and $b$ do not add up to 10.
On
Simple, split the trinomial into a binomial and monomial $(a+3b)+2$. Realize the power of $a+3b$ that contains $a^2b^6$, is the 8th power as exponents add to 8. The powers themselves, force it to happen in the 8 choose 6 coefficient, which is normally equal to 28. If b is raised to the 6th power so was the 3, producing a 729 multiplier. Lastly, the 8th power of $a+3b$ meets up with the second power of 2 ( aka 4). Multiplying these we get 81,648 .
Have a nice day.
One step at a time, $\;\big(a+(3b+2)\big)^{10} = \sum_{k=0}^{10} \binom{10}{k}a^k(3b+2)^{10-k}\,$. The only term containing $\,a^2\,$ is $\,\binom{10}{2}a^2(3b+2)^8\,$. Now expand $\,(3b+2)^8=\sum_{k=0}^{8}\binom{8}{k}(3b)^k2^{8-k}\,$, and note that the only term containing $\,b^6\,$ is $\,\binom{8}{6}(3b)^62^2\,$. Next, put the two together.
Or, use the multinomial expansion directly, where the coefficient of $a^2(3b)^62^2$ is $\binom{10}{\,2, \,6, \,2\,}\,$.