how many sequences of the length 10 and elements $\{a,b,c\}$? How many sequences of the length $10$, with elements from $\{a, b, c\}$ have exactly four $“a”$ and at least three $“b”$?
Can anyone tell me if I am right?
$$\binom{10}{4}\binom{6}{3} +\binom{10}{4}\binom{6}{4}+ \binom{10}{4}\binom{6}{5} +\binom{10}{4}\binom{6}{6} $$

You're given that there are exactly four $a$ and at least three $b$. Thus, you need only choose three more elements from the set $\left\{b,c\right\}$ to get the total number of multisets satisfying these constraints, which is these four:


*

*$\left\{a,a,a,a,b,b,b,b,b,b\right\}$

*$\left\{a,a,a,a,b,b,b,b,b,c\right\}$

*$\left\{a,a,a,a,b,b,b,b,c,c\right\}$

*$\left\{a,a,a,a,b,b,b,c,c,c\right\}$


Adding the number of ways to permute each of these gives the total number of sequences possible, given these constraints. In each case, we can start with $10!$, and then divide out the $4!$ ways to permute the $a$'s that make no difference. This gives us a starting point of $\frac{10!}{4!} = \binom{10}4\cdot6!$ for each case.
Now, to finish the job, we note that in (1) we have to divide out the $6!$ ways to permute the $b$'s that make no difference, giving only $\binom{10}4$ here (or, equivalently, $\binom{10}4\binom66$, as you wrote). For case (2), we divide out $5!$, giving $\binom{10}4 \cdot 6 = \binom{10}4\binom65$. Similarly, we will have to divide out $2!$ and $4!$ for (3), giving $\binom{10}4\binom64$, and we will divide out $\left(3!\right)^2$ for (4), giving $\binom{10}4\binom63$.
Adding up these possibile permutations for all the multisets gives $\binom{10}4\binom63 + \binom{10}4\binom64 + \binom{10}4\binom65 + \binom{10}4\binom66$, so you are indeed correct. However, I felt that a (hopefully) intuitive explanation of why that is so might be of use to you.
