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

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==Problem==
 
==Problem==
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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.)
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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>
  
 
==Solution==
 
==Solution==
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We can separate each rearrangement into two parts: the vowels and the consonants. There are <math> 2 </math> possibilities for the first value and <math> 1 </math> for the remaining one, for a total of <math> 2\cdot1=2 </math> possible orderings of the vowels. There are similarly a total of <math>5!=120 </math> possible orderings of the consonants. However, since both T's are indistinguishable, we must divide this total by <math>2!=2</math>. Thus, the actual number of total orderings of consonants is <math>120 \div 2=60</math>. Thus In total, there are <math> 2\cdot60=120 </math> possible rearrangements, <math>\boxed{\text{(B)}}</math>.
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We consider the vowels and consonants separately. There are <math>2</math> vowels (<math>O</math> and <math>E</math>), giving <math>2! = 2</math> choices for the first two letters; similarly, there are <math>5</math> consonants (<math>C</math>, <math>N</math>, <math>S</math>, and two <math>T</math>s), which would give <math>5! = 120</math> possible choices for letters <math>3</math> to <math>7</math>, except that since the two <math>T</math>s are indistinguishable, this actually counts each order exactly twice. Therefore the number of possible orderings of the consonants is <math>\frac{120}{2} = 60</math>, giving a total of <math>2 \cdot 60 = \boxed{\text{(B)} \ 120}</math> possible rearrangements.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=10|num-a=12}}
 
{{AHSME box|year=1985|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:40, 19 March 2024

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.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AHSME Problems and Solutions

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