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

(Created page with "==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....")
 
(→‎Solution: changed answer)
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==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 <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. In total, there are <math> 2\cdot120=240 </math> possible rearrangements, <math> \boxed{\text{C}} </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 <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>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=10|num-a=12}}
 
{{AHSME box|year=1985|num-b=10|num-a=12}}

Revision as of 20:18, 9 February 2012

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 (Problems • Answer Key • Resources)
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
All AHSME Problems and Solutions
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