Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | Note that a <math>7</math> digit increasing integer is determined once we select a set of <math>7</math> digits. To determine the number of sets of <math>7</math> digits, consider <math>9</math> urns labeled <math>1,2,\cdots,9</math> (note that <math>0</math> is not a permissible digit); then we wish to drop <math>7</math> balls into these urns. Using the ball-and-urn argument, having <math>9</math> urns is equivalent to <math>8</math> dividers, and there are <math>{8 + 7 \choose 7} = {15 \choose 7} = 6435 \equiv \boxed{435} \pmod{1000}</math>. | + | Note that a <math>7</math> digit increasing integer is determined once we select a set of <math>7</math> digits. To determine the number of sets of <math>7</math> digits, consider <math>9</math> urns labeled <math>1,2,\cdots,9</math> (note that <math>0</math> is not a permissible digit); then we wish to drop <math>7</math> balls into these urns. Using the ball-and-urn argument, having <math>9</math> urns is equivalent to <math>8</math> dividers, and there are <math>{8 + 7 \choose 7} = {15 \choose 7} = 6435 \equiv \boxed{435} \pmod{1000}</math>. |
==See Also== | ==See Also== | ||
Latest revision as of 09:16, 29 November 2019
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
Note that a digit increasing integer is determined once we select a set of digits. To determine the number of sets of digits, consider 
urns labeled 
(note that 
is not a permissible digit); then we wish to drop balls into these urns. Using the ball-and-urn argument, having urns is equivalent to 
dividers, and there are 
.
See Also
| Mock AIME 3 Pre 2005 (Problems, Source) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
