Princeton GMAT Math Bible------Permutation

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I posted six questions from Princeton GMAT Math Bible.My thread was blocked because I don't know I should post one question at one time.

I don't understand the following permutation question from Princeton GMAT Math Bible,although there is step by step solution on the book.


8. In how many ways can five people sit around a round table?
Key: 24
Explantation:
Notice: the line permutation is , the circle permutation is . The number of elements of circle permutation is one less than that of line permutation.

Example 1: Put two keys in a key ring which already has five keys. What is the probability that the two new keys will be together?
Way1: first put one new key--A, then put another new key--B: there are 6 keys, and only two places that A and B will be together: 2/6.
Way2: the line permutation is , the circle permutation is

Example 2: There are five points on a circle. One of them is a red point. Five men—A, B, C, D, and E stand on each point. If A must step on the red point, how many ways can be?
The permutation of 4 points excluding A is , after that, there are 4 optional places for point A. therefore, the answer is .

Example 3: there are 6 dishes in a circle—one blue, and five white dishes. There five types of nuts—N, R, G, M, F. if N or R must be in blue dish, and each dish has one nut. How many ways can be?

There is only one way to permutate one blue and five white dishes. We only need consider the permutation of nuts. There is two options for blue dish. The permutation for the nuts in five white dishes is .



Anybody can give me an easy explaination?
Thnx a lot. :oops:
Source: — Problem Solving |

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by dmateer25 » Tue Dec 09, 2008 7:00 am
The number is (n-1)! instead of the usual factorial n! since all cyclic permutations of objects are equivalent because the circle can be rotated.
Check out this website:

https://mathworld.wolfram.com/CircularPermutation.html

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Another question

by amyyoung999 » Tue Dec 09, 2008 7:23 am
Thanks again.
It's very clear.
Can you explain this question for me?



71. Put four letters into four enelopes, what is the probability that only one letter is matching the address on the envelope?

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by reneerenee » Wed Dec 10, 2008 3:02 pm
i think there are 8 ways to put 4 letters in the envelopes to make only one be the right letter. there are 4 ways to choose the right letter, and then only two ways to scramble the remaining 3 ltters so that none of them match.

then 8/24 = 1/3 is the right answer becoause there are 24 ways to do the envelopes total.

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