# Making $10$-digit numbers, using only three digits, and with an additional constraint

Having only digits $1,2,3$. How many $10$-digit numbers can you make with these digits such that you do not use 1 at all? ($2^{10}$?)

How many $10$-digit numbers can you make with these digits provided you use $1$ twice? ($2^8$?)

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Why do you think it is $2^{10}? What led you to guess that (what if all three were allowed)? In the second problem, is it exactly two 1's or at least two 1's (your answer will obviously be different depending)? – Mitch Apr 10 '11 at 14:43 ## 2 Answers Your answer to the first question is correct. In the second question you have to think about places where two 1's will stand. You can choose this in$\binom{10}{2}\$ ways.

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Hint: For the second question (the first seems to be clear). Did you take into account that the two 1's can be at different positions?

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