Difference between revisions of "1985 AHSME Problems/Problem 11"

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==Problem==
 
==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.)
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How many distinguishable rearrangements of the letters in <math>CONTEST</math> have both the vowels first? (For instance, <math>OETCNST</math> is one such arrangement but <math>OTETSNC</math> is not.)
  
 
<math> \mathrm{(A)\ } 60 \qquad \mathrm{(B) \ }120 \qquad \mathrm{(C) \  } 240 \qquad \mathrm{(D) \  } 720 \qquad \mathrm{(E) \  }2520 </math>
 
<math> \mathrm{(A)\ } 60 \qquad \mathrm{(B) \ }120 \qquad \mathrm{(C) \  } 240 \qquad \mathrm{(D) \  } 720 \qquad \mathrm{(E) \  }2520 </math>

Revision as of 23:50, 2 April 2018

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 can separate each rearrangement into two parts: the vowels and the consonants. There are $2$ possibilities for the first value and $1$ for the remaining one, for a total of $2\cdot1=2$ possible orderings of the vowels. There are $5$ possibilities for the first consonant, $4$ for the second, $3$ for the third, $2$ for the second, and $1$ for the first, for a total of $5\cdot4\cdot3\cdot2\cdot1=120$ possible orderings of the consonants. However, since both T's are indistinguishable, we must divide this total by $2!=2$. Thus, the actual number of total orderings of consonants is $120/2=60$ In total, there are $2\cdot60=120$ possible rearrangements, $\fbox{\text{(B)}}$.

See Also

1985 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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
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