$x_1,x_2,x_3,x_4,x_6,x_7$ be 7 digits each of whom can take 1 or 2 or 3. We need to collect 11 from their sum.
Let me represent value=1 as c, value=2 as $c^2$ , value= 3 as $c^3$.
$(c^1 + c^2 + c^3)$ shows all the possible contribution of one digit.
$(c^1+c^2+c^3)(c^1+c^2+c^3)(c^1+c^2+c^3)(c^1+c^2+c^3)(c^1+c^2+c^3)(c^1+c^2+c^3)(c^1+c^2+c^3)$
These are the contribution of 7 digits and all we need to do is to find only those combinations which can make degree of c=11
This would give us the ways in which we can collect 11.
$(c^1+c^2+c^3)^7 = (c(c^0+c^1+c^2))^7$
$=c^7 (c^0+c^1+c^2)^7 ---->$ from here we need to get coefficient of $ [c^{11}]$
$=(1+c^1+c^2)^7 ----> [c^4]$
$=( (1+c^1)+c^2)^7 ----> [c^4]$
In how many ways can we get $c^4$ from here:
1) $\binom{7}{2} (1+c^1)^5 (c^2)^2$
$(1+c^1)^5 = \binom{5}{0} (c^1)^0$
So, $\binom{7}{2} * \binom{5}{0}=21$ is the coefficient of $c^4$
2)
$\binom{7}{1} (1+c^1)^6 (c^2)^1$
$(1+c^1)^6 = \binom{6}{2} (c^1)^2$
So, $\binom{7}{1} * \binom{6}{2} = 105$ is the coefficient of $c^4$
3)
$\binom{7}{0} (1+c^1)^7 (c^2)^0$
$(1+c^1)^7 = \binom{7}{4}(c^1)^4$
So, $\binom{7}{0} * \binom{7}{4}=35$ is the coefficient of $c^4$
Total sum is 21+105+35=161