## Precalculus (6th Edition) Blitzer

$210$
$^{10}C_{4}=\dfrac{10! }{(10-4)!}$ We know that $n!=1 \cdot 2 \cdot 3 .....(n-1)n$ Thus, we have $\dfrac{10! }{(10-4)!}$ $= \dfrac{10 \cdot 9\cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \times 2 \cdot 1 }$ $=210$