document.write( "Question 1189433: What is the probability that a random arrangement of the letters in the word `SEVEN' will have both E's next to each other?
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Algebra.Com's Answer #820816 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "The \"6/15\" answer from the other tutor would almost certainly not be accepted on an exam, since the fraction in not in simplest form. The correct answer to the problem is 2/5.

\n" ); document.write( "And the method shown for solving the problem is unnecessarily long and complicated.

\n" ); document.write( "Here are a couple of more direct ways to solve the problem.

\n" ); document.write( "(1) Standard combinatorics....

\n" ); document.write( "The number of arrangements of 5 objects is 5!=120.

\n" ); document.write( "For the number of arrangements in which the two E's are together, consider them as one unit. Then there are 4 objects to be arranged, in 4!=24 ways; and within the group of the two E's the two can be arranged in either of 2 ways. So the number of arrangements with the 2 E's together is (4!)(2)=48.

\n" ); document.write( "And so the probability that the two E's will be next to each other is 48/120=2/5.

\n" ); document.write( "(2) Using logic to greatly simplify the required calculations....

\n" ); document.write( "The number of ways we can choose 2 of the 5 places for the two E's is C(5,2)=10.
\n" ); document.write( "The number of ways we can choose 2 places in the line that are next to each other is 4.
\n" ); document.write( "The probability that the two E's will be next to each other is 4/10 = 2/5.

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