Date: 07/12/99 at 01:42:03 From: Katia Adge Subject: Reducing the number of permutations using the probability of occurrences How can I reduce the number of arrangements of the word ARRANGEMENT by using the probability of an occurrence? In this example, knowing that this word has 11 letters, 2 R, 2 A, 2 E, and 2 N, the number of permutations of the word ARRANGEMENT is 11!/2!2!2!2! Knowing that 2 N cannot be found next to each other, and the same for 2 E and 2 A, do you know how can I use this information to reduce the number of arrangements of this word? Thank you for answering my question.
Date: 07/12/99 at 07:15:47 From: Doctor Anthony Subject: Re: Reducing the number of permutations using the probability of occurrences If you want the number of arrangements when the 2 N's, 2 E's and 2 A's are not together, you first find the number of arrangements with one or more together and subtract this result from 11!/(2!2!2!2!). To find the number of arrangements with 2 N's together, we join them together as one object, giving a total now of 10 letters. Number of arrangements with N's together = 10!/(2!2!2!). Similarly with two E's = 10!/(2!2!2!) and 2 A's = 10!/(2!2!2!). Now consider two N's together and 2 E's together. We now have 9 objects and these can be arranged in 9!/(2!2!) ways. Similarly with E's and A's together and A's and N's together. Finally with all three letters joined as pairs the number of arrangements is 8!/2! Now using the notation n(N), n(E), n(E and A) to mean the number of arrangements with these letters joined together in pairs, we get from the inclusion-exclusion principle the formula n(N or E or A) = n(N) +n(E) +n(A) -n(N and E) -n(E and A) -n(A and N) + n(N and E and A) = 3 x 10!/(2!2!2!) - 3 x 9!/(2!2!) + 8!/2! = 1108800 And 11!/(2!2!2!2!) = 2494800. So the number without the N's, E's or A's together is 2494800 - 1108800 = 1386000 ways. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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