How many eight-digit positive integers can be written using only the digits 1 and 2 and have exactly five 1’s and three 2's as digits?

_ _ _ _ _ _ _ _     8 digits

"Exactly five 1's and three 2's"

So, let's just write any 8 digits that work:

1 1 1 1 1 1 2 2 2   

Now, we want all the possible permutations without repetition. 

I.e. we want: 1 1 1 1 1 2 1 2 2 and so on.

How many of these?
It is the same idea as, how many words can be made from the word "ACCURACY"

We have 8 digits, 5 of them are 1's (same thing) and 3 of them are 2's (same thing)
So:     \(\frac{8!}{5!*2!}\)  =168 .

(this is saying, arrange the 8 digits as you want, then we divide by 5! as the 1's are repeated, and divide by 2! as the 2's are repeated.) 

I hope this helps!