Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 2"
(should I edit the question? N is less than 1000.) |
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==Solution== | ==Solution== | ||
| − | {{ | + | Since the digits must be in increasing order, they must all be non-zero. We choose 7 digits out of 9, and when we do, they have only one order, so we choose them regardless of order, or <math>\binom{9}{7}=\binom{9}{9-7}=\dfrac{9\cdot 8}{2}=\boxed{036}</math>. |
==See also== | ==See also== | ||
| + | |||
| + | [[Category:Intermediate Combinatorics Problems]] | ||
Revision as of 18:46, 19 June 2008
Problem
Let 
denote the number of 
digit positive integers have the property that their digits are in increasing order. Determine the remainder obtained when is divided by 
. (Repeated digits are allowed.)
Solution
Since the digits must be in increasing order, they must all be non-zero. We choose 7 digits out of 9, and when we do, they have only one order, so we choose them regardless of order, or 
.
