SOLUTION: 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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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?
Found 3 solutions by math_tutor2020, greenestamps, Alan3354:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Edit: I corrected the final answer to have it fully reduced. I appreciate @greenestamps pointing out the error.

If we could tell the two "E"s apart, then there would be 5! = 5*4*3*2*1 = 120 permutations of the word SEVEN since there are 5 letters.

However, we cannot tell the "E"s apart. This means we've double-counted when computing that figure of 120. So we must cut that figure in half to get 60.

There are 60 ways to arrange the letters of SEVEN where the "E"s may or may not be together.

Let's take the "E"s out and replace it with some other letter. I'll use W.
So we go from the word SEVEN to the "word" SWVN

There are 4 distinct letters here, so there are 4! = 4*3*2*1 = 24 permutations. Anywhere you see a W, replace it with EE

A permutation like WSVN is really EESVN
A permutation like SVWN is really SVEEN
and so on.

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We found that there are: Divide the two values to get: 24/60 = (12*2)/(12*5) = 2/5


Answer: 2/5
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


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.

And the method shown for solving the problem is unnecessarily long and complicated.

Here are a couple of more direct ways to solve the problem.

(1) Standard combinatorics....

The number of arrangements of 5 objects is 5!=120.

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.

And so the probability that the two E's will be next to each other is 48/120=2/5.

(2) Using logic to greatly simplify the required calculations....

The number of ways we can choose 2 of the 5 places for the two E's is C(5,2)=10.
The number of ways we can choose 2 places in the line that are next to each other is 4.
The probability that the two E's will be next to each other is 4/10 = 2/5.
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
both E's next to each other?
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Both.
As opposed to only one of them?
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What is the probability that a random arrangement of the letters in the word `SEVEN' will have the E's next to each other?
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