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In how many ways can the letters of the word NUMBER be arranged? In how many ways is the letter N somewhere to the left of U?

Why is the answer to the second question exactly half of the answer to the first?

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  • $\begingroup$ Not quite sure this is the right answer, but U is either before or after N, and I would imagine there is a symmetry to it, so the probability of it being before is the same as the probability of it being after, since you can always take mirror images of the combination so the probability is 1/2. $\endgroup$ – H_Hassan Dec 20 '16 at 20:51
  • $\begingroup$ Please edit your question and make it clearer. It is unclear. $\endgroup$ – MrAP Dec 20 '16 at 20:51
  • $\begingroup$ Done, i think it's better now $\endgroup$ – TripleA Dec 20 '16 at 20:54
  • $\begingroup$ Why is it that A precedes B in exactly half of the sequences "AB", "BA"? $\endgroup$ – Wildcard Dec 21 '16 at 1:44
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Because every letter in NUMBER is distinct, for every permutation that N comes before U, there must be a mirrored permutation where U comes before N. This relationship is also bijective because the permutations with N before U are distinct, so we see that when N is before U, there is a complementary permutation with the N and U swapped, so therefore, we get a probability of $\frac{1}{2}$ for the permutations where N is before U.

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Expanding on H_Hassan's answer: consider all permutations of NUMBER. Partition them into two classes: the words where U is to the right of N, and where U is to the left of N. We can construct a bijection between the two classes. For any word in the first class, consider its reversal e.g. the reversal of NUMBER is REBMUN. Then you have a word in the second class. This is a bijection, because if two words have the same reversal, they must be the same word, and if you have any word in the second class, then its reversal is a corresponding word in the first class whose reversal is the word in the second class. Hence, since these sets are bijective, they have the same size.

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Suppose N and U are identical letters. Then $6!/2!=360$ arrangements are possible.

Actually, N and U are distinct. So, we can swap positions of N and U in each of the $360$ arrangements and in one case, N will be left to U and another case, N will be right to U. So, equal probability for both.

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The word "NUMBER" is 6 lettered.

For the first question, we want all the possible words that can be formed from the letters of this word, i.e., all the possible permutations of the letters of this word which simply comes out to be 6! that is equal to 720.

To understand the answer to your actual question we have to understand that there are two possible cases. One in which N comes before U and the other in which N comes after U. We are concerned about the words in which the letter N comes before U. So obviously, N cannot be at the 6th position since in that case the E cannot be to the right of N. The number of permutations comes out to be (5!*2)+(4!*5) which is equal to 360. A similar argument can be made for the second case or you can just subtract 360 from 720 to get the number of words for the second case and the number of words in the second case too comes out to be 360.

In conclusion the number of total number of words possible from the letters of "NUMBER" is 720. The words can be categorized into two types : one in which N is before U and the other in which N is after U. Due to symmetricity of both the cases, the both are each equal to 360 which is half of 720.

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