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how many ways can the letters in ARRANGEMENT can be arranged

a) In how many different ways can the letters in the word ARRANGEMENTS be arranged?

b) Find the probability that an arrangement chosen at random begins with the letters EE

Some Guy
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darshanie M
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  • Part (a) is identical to the question you asked earlier today. Your earlier question will get "bumped" up if you take the time to edit your earlier question to include what you do not understand from the posted answers. Likewise, you can add part (b) to that post. – amWhy Nov 13 '12 at 19:45
  • @amWhy: Well, almost identical. This one has an S on the end. Still, darshanie, I recommend that you follow amWhy's recommendation. Edit your earlier question and delete this one. – Cameron Buie Nov 13 '12 at 20:27

1 Answers1

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Part a) was asked and answered very recently.

b) In general, we need to know something about the process by which the randomization took place. We assume that the letters were put in a bag, and the bag was shaken. We picked a first letter, then a second.

The probability that the first letter picked is E is $\dfrac{2}{11}$. Given that the first letter picked was E, the probability that the second was E is $\dfrac{1}{10}$. So the required probability is $\dfrac{2}{11}\cdot\dfrac{1}{10}$.

Alternately we can persuade ourselves that the arrangements counted in the recent post are all equally likely. Then we can count the "favourable" arrangements, that is, the arrangements that begin with EE. To count, remove the two E's. We have $9$ letters left, with some repetitions. An argument similar to the one used to count the arrangements of ARRANGEMENT can be used to show that the remaining $9$ letters can be arranged in $$\binom{9}{2}\binom{7}{2}\binom{5}{2}3!$$ ways. Now divide by the answer to the previous question.

Another way: Let us colour the duplicated letters in ARRANGEMENT, to make them all distinct. So we have a red E and a blue E, and so on.

There are then $11!$ words, all equally likely. To count the favourables, the first letter can be chosen in $2$ ways (red E or blue E). For each such choice, the second letter can be chosen in only $1$ way. Now the rest of the letters can be arranged in $9!$ ways. So our probability is $$\frac{(2)(1)(9!)}{11!},$$ which can be greatly simplified.

Community
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André Nicolas
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  • Answering part(b) here, I'm afraid, rewards the OP for not working with his/her previous post, by editing to ask for clarification and/or add part (b). – amWhy Nov 13 '12 at 19:49
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    @amWhy: I agree that part a) should not have been re-asked, or if it was, there should have been a link to the earlier question. But asking a new question is I think in general much better than editing an old one, for editing can make earlier answers irrelevant or incomplete. – André Nicolas Nov 13 '12 at 20:03