Forming a committee with three restrictions

Question

Suppose that in the US Supreme Court a committee of seven politicians is chosen from five republicans, ten democrats and eight independents. How many different ways can the committee be chosen if the committee must include exactly one republican, at least three democrats and at least one independent?

I first considered the number of ways of forming a committee with at least three democrats and thought this was equal to $$ C(10,3)\times C(13,4)+C(10,4)\times C(13,3)+C(10,5)\times C(13,2)=165,516. $$ However the answer in my book is $73,080$. I am yet to consider the other restrictions and yet my answer is too large. Where am I double counting?

Answer 1

Apart from Democrats, you have lumped all together, which is patently wrong, and will overcount by violating constraints for Republicans.

Start by choosing $1$ Republican from $5$, and follow the restrictions for the other two parties to get the right answer.

$\binom51\left[\binom{10}3\binom83 +\binom{10}4\binom82 +\binom{10}5\binom81\right]$