Unable to think about 10-combinations of a multiset

Question

I am trying assignments of an Institute in which I don't study and i could not think about this problem.

Problem is -> Determine the number of 10-combinations of multisets S= { 3.a, 4.b, 5.c}

I think the answer should be number of solutions of equation $ x_1+ x_2+x_3$ =10 such that $0\leq x_1\leq 3 $ , $0 \leq x_2 \leq 4 $ , $ 0\leq x_3 \leq 5 $ .

But i am unable to think how to find solution of this equation under such constraints. Can somebody please help.

Answer 1

[EDIT: Btw, your interpretation as ways to sum to $10$ is completely correct :)]

I will just give two (and a half) other ways to work it out.

Since $|S|=12$ and we want to count $10$-combinations, it is simpler to count $2$-combinations instead (i.e. choose the elements not to go in the $10$-combination). There are $6$ of these: $$ \{a,a\}, \{b,b\}, \{c,c\}, \{a,b\}, \{a,c\}, \{b, c\} $$ These can be counted systematically by noting that we must choose either two of the same or two distinct elements, and then use basic combinatorics.

We can also note that $a,b,c$ all have multiplicity $\ge2$ in $S$. So the $2$-combinations of $S$ are the same as the $2$-combinations of the set $\{a,b,c\}$ with unlimited repetition. By stars and bars, that is $\binom{2+3-1}{2}=6$.

If we want to use a direct formula, we have to deal with a bunch of cases through inclusion-exclusion. With this method (as seen here), we get: $$ \binom{10+3-1}{2} - \Bigg[ \binom{6+3-1}{2} + \binom{5+3-1}{2} + \binom{4+3-1}{2} \Bigg] \\ + \Bigg[ \binom{1+3-1}{2} + \binom{0+3-1}{2} + 0 \Bigg] - 0 = 6 $$ (If you follow the link, note that the $56$ in their example should be $\binom{8}{2}=28$, and the correct answer is $9$).

Answer 2

I'm interpreting the problem in the sense indicated in the sentence "I think the answer should be $\ldots$".

As $0\leq x_3\leq5$ we must have $x_1+x_2\geq5$. Draw a figure of the $(x_1,x_2)$-plane, and you shall immediately see that there are exactly $6$ lattice points in the $[0,3]\times[0,4]$ rectangle satisfying this condition, namely $(1,4)$, $(2,3)$, $(2,4)$, $(3,2)$, $(3,3)$, and $(3,4)$. Since $(x_1,x_2)$ determines $x_3$ via $x_3=10-(x_1+x_2)$ it follows that the given problem has $6$ admissible solutions $(x_1,x_2,x_3)$.