Sunday

April 20, 2014

April 20, 2014

Posted by **Sana** on Sunday, October 15, 2006 at 10:37am.

I'm not entirely sure what you question is asking, but I'll take a guess.

You have n digits and k 1's with k<n

There are n choose k ways to arrange the 1's in your number.

You now have n-k places to choose the other 9 numbers. There are two cases to consider: when the fist digit is 1 and when it isn't. If it's not 1 then there are only 8 digits you can use, since a number wouldn't have 0 in the lead position. If it is a 1 then there are 9 possible choices for each of the n-k position or 9

I see someone else posted this question too and they said leading 0's are acceptable, so you can disregard part of the previous post if you have the same quesiton as them.

There are n choose k ways to arrange the 1's and 9

n choose k * 9

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