1. **Problem Statement:** Find the number of integer solutions to the equation $$a+b+c+d+e+f+g=10$$ where each variable $$a,b,c,d,e,f,g$$ satisfies $$1 \leq a,b,c,d,e,f,g \leq 3$$. 2. **Reformulate the problem:** Since each variable is at least 1, let $$a'=a-1, b'=b-1, \ldots, g'=g-1$$. Then each $$a',b',\ldots,g'$$ lies between 0 and 2, and the equation becomes: $$a'+b'+c'+d'+e'+f'+g' = 10 - 7 = 3$$ 3. **Constraints:** Each $$a',b',\ldots,g'$$ satisfies $$0 \leq a',b',\ldots,g' \leq 2$$. 4. **Counting solutions without upper bound:** The number of nonnegative integer solutions to $$a'+b'+c'+d'+e'+f'+g' = 3$$ without upper bounds is given by the stars and bars theorem: $$\binom{3+7-1}{7-1} = \binom{9}{6} = 84$$ 5. **Applying upper bound constraints:** We must exclude solutions where any variable exceeds 2. 6. **Use Inclusion-Exclusion Principle:** - Let $$A_i$$ be the set of solutions where the $$i^{th}$$ variable $$> 2$$. - Since the sum is 3, a variable $$> 2$$ means it is at least 3. 7. **Counting solutions with one variable $$\geq 3$$:** Set $$a'_i = b_i + 3$$ where $$b_i \geq 0$$. Then the equation becomes: $$b_i + \sum_{j \neq i} a'_j = 3 - 3 = 0$$ Number of solutions: $$\binom{0+7-1}{7-1} = \binom{6}{6} = 1$$ 8. **Counting solutions with two or more variables $$\geq 3$$:** Impossible since sum would be at least 6. 9. **Inclusion-Exclusion formula:** $$\text{Valid solutions} = 84 - \binom{7}{1} \times 1 = 84 - 7 = 77$$ **Final answer:** There are **77** integer solutions to the equation under the given constraints.