Find the number of integer solutions to $x_1 + x_2 + x_3 + x_4 = 25$ with $ 1 \leq x_1 \leq 6, 2 \leq x_2 \leq 8, 0 \leq x_3 \leq 8, 5 \leq x_4 \leq 9.$
Firstly, I defined $y_i = x_i - lower bound$ so for example $y_1 = x_1 - 1$ This gives $y_1 + y_2 + y_3 + y_4 = 17$ with $$0 \leq y_1 \leq 5, 0 \leq y_2 \leq 6, 0 \leq y_3 \leq 8, 0 \leq y_4 \leq 4$$ Next I applied Inclusion-Exclusion Principle;
Total number of ways = $17+3 \choose 3$
Number of ways with just $y_i$ breaking upper bound;
$y_1;$$11+3 \choose 3$
$y_2;$ $10+3 \choose 3$
$y_3;$ $8+3 \choose 3$
$y_4;$ $12+3 \choose 3$
Similarly I calculated the number of ways with $y_i$ and $y_j$ breaking upper bound at the same time. And since there is no way in which 3 of them break their upper bounds simultaneously I was left with ;
$${20 \choose 3} - {14 \choose 3} - {13 \choose 3} - {11 \choose 3} - {15 \choose 3} + {7\choose 3} + {5 \choose 3} + { 9 \choose 3} + {4 \choose 3} + {8 \choose 3} + {6 \choose 3} $$
as my final answer.
Is this solution correct and is there a nicer way of solving it/ any insights about this type of problem?
Note, this problem is from 'An Introduction to Combinatorics and Graph Theory'
As noted in the comments, your solution is correct. The approach is described in Balls In Bins With Limited Capacity.
To solve the problem using generating functions, write $x+x^2+\cdots+x^6=x\frac{1-x^6}{1-x}$ for $x_1$, $x^2+x^3+\cdots+x^8=x^2\frac{1-x^7}{1-x}$ for $x_2$, $1+x+\cdots+x^8=\frac{1-x^9}{1-x}$ for $x_3$ and $x^5+x^6+\cdots+x^9=x^5\frac{1-x^5}{1-x}$ for $x_4$. The product is
$$ x^8\frac{\left(1-x^6\right)\left(1-x^7\right)\left(1-x^9\right)\left(1-x^5\right)}{(1-x)^4}\;, $$
and the number of solutions to $x_1+x_2+x_3+x_4=25$ is the coefficient of $x^{25}$ in this generating function. With the expansion
$$ \frac1{(1-x)^4}=\sum_{k=0}^\infty\binom{k+3}3x^k\;, $$
you can see how multiplying out the numerator and matching each resulting term with the missing powers of $x$ from the denominator yields exactly the alternating inclusion–exclusion sum of binomial coefficients that you calculated.