Difference between revisions of "2006 Alabama ARML TST Problems/Problem 9"
(New page: ==Problem== Amanda ordered a dozen donuts. She said she wanted only chocolate, glazed, and powdered donuts, and at least one of each kind. Let <math>a</math>, <math>b</math>, and <math>c</...) |
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There are 11 spaces between the D's, so there are <math>\binom{11}{2}</math> ways to set the dividers. Therefore, there are <math>\boxed{55}</math> ways to get three kinds of donuts totaling 12. | There are 11 spaces between the D's, so there are <math>\binom{11}{2}</math> ways to set the dividers. Therefore, there are <math>\boxed{55}</math> ways to get three kinds of donuts totaling 12. | ||
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| + | ==See Also== | ||
| + | {{ARML box|year=2006|state=Alabama|num-b=5|num-a=7}} | ||
Revision as of 07:29, 15 April 2008
Problem
Amanda ordered a dozen donuts. She said she wanted only chocolate, glazed, and powdered donuts, and at least one of each kind. Let 
, 
, and 
be the number of chocolate, glazed, and powdered donuts she wound up with. Find the number of possible ordered triples 
.
Solution
She has a dozen unknown donuts: D, D, D, D, D, D, D, D, D, D, D, D. We can set dividers between the D's and come up with some chocolate, glazed, and powdered donuts using the following rules:
We set two dividers between the D's so that they split the D's into three groups. The left group is chocolate, the middle group is glazed, and the right group is powdered.
There are 11 spaces between the D's, so there are 
ways to set the dividers. Therefore, there are 
ways to get three kinds of donuts totaling 12.
See Also
| 2006 Alabama ARML TST (Problems) | ||
| Preceded by: Problem 5 |
Followed by: Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
