## Precalculus (6th Edition) Blitzer

$\dfrac{50}{5005}$
There are a total of $^{15}C_{6}$ possible ways to choose $6$ numbers for the winning lottery combination and this can be our denominator. We know that $n!=1 \cdot 2 \cdot 3 .....(n-1)n$ Thus, we have $\dfrac{50!}{^{15}C_{6}}=\dfrac{50 !}{\dfrac{15! }{(15-6)!}}$ $= \dfrac{50}{5005}$