Difference between revisions of "1989 AHSME Problems/Problem 9"

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==See also==
 
==See also==
 
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Revision as of 12:47, 5 July 2013

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?

$\textrm{(A)}\ 276\qquad\textrm{(B)}\ 300\qquad\textrm{(C)}\ 552\qquad\textrm{(D)}\ 600\qquad\textrm{(E)}\ 15600$

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 $\dbinom{25}{2}=\boxed{\mathrm{(B)\,300}}$.

See also

1989 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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