Difference between revisions of "1989 AHSME Problems/Problem 9"
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==Solution== | ==Solution== | ||
| − | We see that for any combination of two distinct letters other than Z (as the last name will automatically be Z), there is only one possible way to arrange them in alphabetical order, thus the answer is just <math>\dbinom{25}{2}=\boxed{\mathrm{( | + | We see that for any combination of two distinct letters other than Z (as the last name will automatically be Z), there is only one possible way to arrange them in alphabetical order, thus the answer is just <math>\dbinom{25}{2}=\boxed{\mathrm{(B)\,300}}</math>. |
==See also== | ==See also== | ||
{{AHSME box|year=1989|num-b=8|num-a=10}} | {{AHSME box|year=1989|num-b=8|num-a=10}} | ||
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| + | [[Category: Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 06:48, 22 October 2014
Problem
Mr. and Mrs. Zeta want to name their baby Zeta so that its monogram (first, middle, and last initials) will be in alphabetical order with no letter repeated. How many such monograms are possible?
Solution
We see that for any combination of two distinct letters other than Z (as the last name will automatically be Z), there is only one possible way to arrange them in alphabetical order, thus the answer is just 
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See also
| 1989 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
