If a die is rolled 10 times, what is the probability that it lands on '6' exactly 5 times?

My approach:

There $\displaystyle {10 \choose 5}$ ways for the number '6' to turn up 5 times in 10 throws.

The probability of a '6' appearing after each throw is 1/6.

The probability of any other number turning up after each throw is 5/6.

So I have,

$\displaystyle P = {10 \choose 5} (1/6)^{5} (5/6)^{5}$

Am I thinking about this the right way? Thanks.