I just was just given this question and the answer to it but can't seem to solve it as I think I am missing something…
I've tried using different approaches such as n!/r!(n-r!) but I have a problem knowing how many options i've got when placing each numbers. Please forgive me as i'm not familiar with combinatorics…
Best Answer
How many 6-digit numbers contain exactly three 1s?
The 6-digit number begins with 1
In this case, suppose we have 6 places. The first place fill with 1. The 2 of remanning places must filled with 1 and 3 of them must filled with {0,2,3,4,5,6,7,8,9}. Thus, we have $\binom{5}{2}\times 9^{3}$
The 6-digit number begins with {2,3,4,5,6,7,8,9}
In this case, we have $8\times \binom{5}{3} \times 9^{2}$
At the end, the result is $\binom{5}{2}\times 9^{3} + 8\times \binom{5}{3} \times 9^{2}$