Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 2"
m |
m |
||
| (4 intermediate revisions by 2 users not shown) | |||
| Line 2: | Line 2: | ||
==Solution== | ==Solution== | ||
| − | {{ | + | Equivalently, we need to place 12 indistinguishable balls into 7 distinguishable boxes so that no box contains more than 9 balls. There are <math>{12 + 7 - 1 \choose 7 - 1} = {18 \choose 6} = 18,564</math> ways to place 12 objects into 7 boxes. Of these, 7 place all 12 into a single box. <math>7 \cdot 6 = 42</math> place 11 in one box and 1 in a second. <math>7 \cdot 6 = 42</math> place 10 into one box and 2 into a second. <math>7 \cdot \frac{6\cdot 5}{2} = 105</math> place 10 into one box and 1 into each of two others. Thus, this gives us <math>m = 18564 - 7 - 42 - 42 - 105 = 18368</math> so <math>\star(m) = 1 + 8 + 3 + 6 + 8 = 026</math>. |
| + | ---- | ||
| − | + | *[[Mock AIME 1 2006-2007 Problems/Problem 1 | Previous Problem]] | |
| − | [[Mock AIME 1 2006-2007/Problem 1 | Previous Problem]] | + | |
| − | [[Mock AIME 1 2006-2007/Problem 3 | Next Problem]] | + | *[[Mock AIME 1 2006-2007 Problems/Problem 3 | Next Problem]] |
| − | [[Mock AIME 1 2006-2007]] | + | |
| + | *[[Mock AIME 1 2006-2007]] | ||
| + | |||
| + | [[Category:Intermediate Combinatorics Problems]] | ||
Latest revision as of 14:53, 3 April 2012
Let 
be the sum of the digits of a positive integer 
. 
is the set of positive integers such that for all elements 
in , we have that 
and 
. If 
is the number of elements in , compute 
.
Solution
Equivalently, we need to place 12 indistinguishable balls into 7 distinguishable boxes so that no box contains more than 9 balls. There are 
ways to place 12 objects into 7 boxes. Of these, 7 place all 12 into a single box. 
place 11 in one box and 1 in a second. place 10 into one box and 2 into a second. 
place 10 into one box and 1 into each of two others. Thus, this gives us 
so 
.
