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{(D)\,300}}</math>.
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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}}

Revision as of 15:53, 22 February 2012

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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All AHSME Problems and Solutions
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