1. The problem asks to evaluate the combination $7 C 3$, which represents the number of ways to choose 3 items from 7 without regard to order. 2. The formula for combinations is: $$n C r = \frac{n!}{r!(n-r)!}$$ where $n!$ is the factorial of $n$. 3. Substitute $n=7$ and $r=3$: $$7 C 3 = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!}$$ 4. Calculate factorials: $$7! = 7 \times 6 \times 5 \times 4!$$ So, $$7 C 3 = \frac{7 \times 6 \times 5 \times 4!}{3! \times 4!}$$ 5. Cancel $4!$ from numerator and denominator: $$7 C 3 = \frac{7 \times 6 \times 5 \times \cancel{4!}}{3! \times \cancel{4!}} = \frac{7 \times 6 \times 5}{3!}$$ 6. Calculate $3!$: $$3! = 3 \times 2 \times 1 = 6$$ 7. Substitute and simplify: $$7 C 3 = \frac{7 \times 6 \times 5}{6}$$ 8. Cancel 6 in numerator and denominator: $$7 C 3 = \frac{7 \times \cancel{6} \times 5}{\cancel{6}} = 7 \times 5 = 35$$ 9. Therefore, the value of $7 C 3$ is an integer: $$7 C 3 = 35$$