Difference between revisions of "1985 AHSME Problems/Problem 11"
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==Problem== | ==Problem== | ||
− | 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.) | + | 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== | ||
− | We | + | 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 19:40, 19 March 2024
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 consider the vowels and consonants separately. There are vowels ( and ), giving choices for the first two letters; similarly, there are consonants (, , , and two s), which would give possible choices for letters to , except that since the two s are indistinguishable, this actually counts each order exactly twice. Therefore the number of possible orderings of the consonants is , giving a total of possible rearrangements.
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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