P & C

$\rightarrow$ There are two possible cases

$\underline{Case 1:-}$

Five 1's, One 2's and One 3's

$Number\quad of\quad numbers\quad =\quad \frac { 7! }{ 5! } \quad =\quad 42$

$\underline{Case 2:-}$

Four 1's, three 2's

$Number\quad of\quad numbers\quad =\quad \frac { 7! }{ 4!\quad 3! } \quad =\quad 35$

Total Number Of Solutions = 42+35 = 77

coefficient of x^10 in the expansion of (x+x^2+x^3)^7 is the answer. .

We use generating functions. Clearly, we are looking for the coefficient of $x^10$ in $(x + x^2+x^3)^7$ , which is equivalent to finding the coefficient of $x^3$ in $(1+x+x^2)^7 = (\frac{1-x^3}{1-x})^7$ . This is clearly $\binom{9}{6} - 7*\binom{6}{6} = 77$