1985 AHSME Problems/Problem 11

Problem

How many distinguishable rearrangements of the letters in $CONTEST$ have both the vowels first? (For instance, $OETCNST$ is one such arrangement, but $OTETSNC$ is not.)

$\mathrm{(A)\ } 60 \qquad \mathrm{(B) \ }120 \qquad \mathrm{(C) \ } 240 \qquad \mathrm{(D) \ } 720 \qquad \mathrm{(E) \ }2520$

Solution

We consider the vowels and consonants separately. There are $2$ vowels ( $O$ and $E$ ), giving $2! = 2$ choices for the first two letters; similarly, there are $5$ consonants ( $C$ , $N$ , $S$ , and two $T$ s), which would give $5! = 120$ possible choices for letters $3$ to $7$ , except that since the two $T$ s are indistinguishable, this actually counts each order exactly twice. Therefore the number of possible orderings of the consonants is $\frac{120}{2} = 60$ , giving a total of $2 \cdot 60 = \boxed{\text{(B)} \ 120}$ possible rearrangements.

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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1985 AHSME Problems/Problem 11

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