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

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==Solution==
 
==Solution==
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 <math> 5 </math> possibilities for the first consonant, <math> 4 </math> for the second, <math> 3 </math> for the third, <math> 2 </math> for the second, and <math> 1 </math> for the first, for a total of <math> 5\cdot4\cdot3\cdot2\cdot1=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>120/2=60</math> In total, there are <math> 2\cdot60=120 </math> possible rearrangements, <math>\fbox{\text{(B)}}</math>.
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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 \cdot1=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>\fbox{\text{(B)}}</math>.
  
 
==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}}

Revision as of 23:51, 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 similarly a total of $5 \cdot1=120$ 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 $120 \div 2=60$. Thus 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
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