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?
$$\binom{10}{4}\binom{6}{3} +\binom{10}{4}\binom{6}{4}+ \binom{10}{4}\binom{6}{5} +\binom{10}{4}\binom{6}{6} $$
 A: 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.
A: Yes you are absolutely right but let me show you my work
Since the string has exactly $4~a's$ and atleast $3~b's$, fix them and then we are left with $3$ places and $b,c$ can be filled in them.
Now there are $4$ different cases of filling those $3$ places 
$1.$ With $0~b's$ and $3~c's$ then total number of words are $\frac{10!}{4!3!3!}=4200$
$2.$ With $1~b's$ and $2~c's$ then total number of words are $\frac{10!}{4!4!2!}=3150$
$3.$ With $2~b's$ and $1~c's$ then total number of words are $\frac{10!}{4!5!1!}=1260$
$4.$ With $3~b's$ and $0~c's$ then total number of words are $\frac{10!}{4!6!0!}=210$
Adding all will give $8620$ , which corresponds to your result.
Although your work is right , but I am still posting this answer because alternative solutions are always best and they help in tackling other problems.
Also I have learnt so much things from this site and still learning, but I din't thought at first that your answer is right and checked it by calculations , and after preparing my result, I got clear with your solution .
So if any new user like me wants to understand this type of questions, then he/she would have two nice answers or more, so that they can understand it completely 
