Find the number of 5-member committees which include at least two Republicans

Question

There are 10 Republicans, 8 Democrats and 2 Independent legislators eligible for committee membership.

How many 5-member committees exist which include at least two Republicans?

My Work :

$C(20,5)-(C(10,5)+C(10,4))$

$(\;All\; 5-member\; committees\;)\;-\;(\;committiees\; with \;no\; Republicans\;+\;committees\; with\; one \;Republican \;)$

Explanation: There are $C(20,5)$ 5-member committees which will be subtracted from sum of two numbers in order to achieve the answer; C(10,5) is number of 5-member committees which include no Republicans (I've only counted Democrats and Independents ($8+2$)then found number of 5-member committees ) and C(10,4) is number of 5-member committees with one Republican (Like before I've excluded Republicans but counted 4-member committees then added one Republican to each subset)

Book's Answer:

It's just said $C(10,2)C(18,3)$

Answer 1

I think that the answer is $$\binom{20}5-\binom{10}5-10\binom{10}4=13,152$$

and that book's answer is wrong. The idea behind book's answer seems to be the following:

First, we choose two Republicans. There are $C(10,2)$ ways to do it. Then we choose three any other members. There are $C(18,3)$ ways to do it. The problem is that then we are double (or even triple, etc) counting many possible choices. For an example, let's say that the Republicans are numbered from $1$ to $10$, and the other ones from $11$ to $20$. If we choose for first $1$ and $2$ and then $3$, $4$ and $12$ it is the same as choosing for first $3$ and $4$ and then $1$, $2$ and $12$.

Note (just in case you don't know): $\binom ab$ is just another way (and more usual, if you ask me) to write $C(a,b)$.

Answer 2

Republicans-$10$,Democrats-$8$,Independent legislator-$2$

There should be at least $2$ Republicans in the $5$ members committee.

So, there may be $2$ or $3$ or $4$ or $5$ republicans in the committee, as there are no restrictions on the numbers of the Democrats & Independent legislator in the committee-so we can treat these two groups as a single one of $8+2=10$ members.Call this new group the D-IL group.

The number of ways of forming such committee is

[($2$ Republicans & $3$ D-IL)or ($3$ Republicans & $2$ D-IL)or ($4$ Republicans & $1$ D-IL)or ($5$ Republicans & $0$ D-IL)]

=($2$ Republicans)$\cdot$ ($3$ D-IL)+($3$ Republicans)$\cdot$ ($2$ D-IL)+($4$ Republicans)$\cdot$ ($1$ D-IL)+($5$ Republicans)$\cdot$($0$ D-IL)

=${10 \choose 2}\cdot {10 \choose 3}+{10 \choose 3}\cdot {10 \choose 2}+{10 \choose 4}\cdot {10 \choose 1}+{10 \choose 5}\cdot {10 \choose 0}=13152.$