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....") |
Freddylukai (talk | contribs) (→Solution: changed answer) |
||
Line 5: | Line 5: | ||
==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. In total, there are <math> 2\ | + | 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 19:18, 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.)
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 possibilities for the first consonant, for the second, for the third, for the second, and for the first, for 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 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 |