What is the sum of ( \frac{10!}{4!\cdot6!} ) and ( \frac{10!}{3!\cdot7!} )?

Correct answer: (330)

Each term has the form \(\frac{n!}{r!(n-r)!}\), which is the combination \(\binom{n}{r}\). Thus the first term is \(\binom{10}{4}=210\), and the second is \(\binom{10}{3}=120\). Recognising this form makes the calculation shorter, although the same result follows by cancelling factorials directly.

Adding the two separately evaluated terms gives \(210+120=330\). Therefore option B is correct. It is important not to add the denominators or combine the numerators before evaluating, because factorial expressions do not distribute over addition in that way. The values 240, 360, and 450 do not result from the correct combination calculations.