Answer
$120$
Work Step by Step
We know that $ n!=1 \cdot 2 \cdot 3 .....(n-1)n $
Thus, we have $\dfrac{10!}{(10-3)!3!}$
$= \dfrac{10 \cdot 9\cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\times 3 \cdot 2 \cdot 1 }$
$=120$
