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 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>\ | + | 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>. |
==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 15:12, 29 November 2020
Problem
How many distinguishable rearrangements of the letters in 
have both the vowels first? (For instance, 
is one such arrangement but 
is not.)
Solution
We can separate each rearrangement into two parts: the vowels and the consonants. There are 
possibilities for the first value and 
for the remaining one, for a total of 
possible orderings of the vowels. There are similarly a total of 
possible orderings of the consonants. However, since both T's are indistinguishable, we must divide this total by 
. Thus, the actual number of total orderings of consonants is 
. Thus In total, there are 
possible rearrangements, 
.
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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