The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is I solved it using the regular combination method taking the two cases as 1,1,1,1,1,2,3 and 1,1,1,1,2,2,2.
But how do you solve it using multinomial theorem?
 A: Here is perhaps an alternative approach: Let us think about how many ways one can form three digit numbers using the digits $1$, $2$, and $3$, such that the result has digital sum 7. The possibilities can be listed exhaustively:
$$133, 313, 331, 233, 323, 332$$
There are a total of six such possibilities.
Next, consider (for reasons perhaps not yet clear) the following:
$$(x^1 + x^2 + x^3)^3 = x^9 + 3x^8 + 6x^7 + 7x^6 + 6x^5 + 3x^4 + x^3$$
Since we only have the digits $1$, $2$, and $3$, the only exponents that apppear within the parentheses are precisely these; i.e., all you see is $x^1$, $x^2$, and $x^3$. Moreover, since we are trying to look at three digit numbers, the expression in parentheses is raised to the third power. Finally, we are wondering about how many of the resulting three digit numbers have digital sum seven; indeed, looking at the expanded form, the coefficient of $x^7$ gives the answer: $6$ (as listed above).
What's going on here? Try tinkering around to see how these exponents interact with one another. As to your particular problem, you can expand $(x^1 + x^2 + x^3)^7$ and check the coefficient of $x^{10}$; as remarked in a comment above, or as you can find yourself using e.g. WolframAlpha, it is $77$.
A: There are only two integer partitions of $10$ of length 7 formed from the digits $1,2,3$:  $\{ 3,2,1,1,1,1,1 \}$ and $\{ 2,2,2,1,1,1,1 \}$.  The number of ways you can form numbers from the first set can be computed using a multinomial:
${7! \over 1! 1! 5!} = 42$
and for the second set is
${7! \over 3! 4!} = 35$
so the total is $77$.
Confirmation using Mathematica:
 Length@Select[Tuples[{1, 2, 3}, 7], Total[#] == 10 &]

(* 
77
*)
A: The number of solutions is the coefficient of $x^{10}$ in $(x+x^2+x^3)^7$ which is the coefficient of $x^3$ in $(1-x^3)^7(1-x)^{-7}$. Only the first two terms : $1$ and $-7x^3$ are of relevance to us since the other terms have higher powers of $x$. If we select $1$, we'd have to select the coefficient of $x^3$ in $(1-x)^{-7}$. I leave it to you to find the other coefficient out. 
The answer is :
$$
\binom{9}{6}-7=77
$$
