Difference between revisions of "2021 AIME I Problems/Problem 4"
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Hence, the answer is <math>\frac{\binom{65}{2} - 96 + 2}{6} = \boxed{331}.</math> | Hence, the answer is <math>\frac{\binom{65}{2} - 96 + 2}{6} = \boxed{331}.</math> | ||
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| + | ==Solution 3== | ||
==See also== | ==See also== | ||
Revision as of 10:56, 12 March 2021
Problem
Find the number of ways 
identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.
Solution 1
Suppose we have 
coin in the first pile. Then 
all work for a total of 
piles. Suppose we have 
coins in the first pile, then 
all work, for a total of 
. Continuing this pattern until 
coins in the first pile, we have the sum 
.
Solution 2
Let the three piles have 
coins respectively. If we disregard order, then we just need to divide by 
at the end.
We know 
. Since are positive integers, there are 
ways from Stars and Bars.
However, we must discard the cases where 
or 
or 
. The three cases are symmetric, so we just take the first case and multiply by 3. We have 
for 32 solutions. Multiplying by 3, we will subtract 96 from our total.
But we undercounted where 
. This is first counted 1 time, then we subtract it 3 times, so we add it back twice. There is clearly only 1 way, for a total of 2.
Hence, the answer is 
Solution 3
See also
| 2021 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
