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1 A 2 3 B

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It follows that there are n! possible permutations of n distinct objects The number of permutations of n distinct objects grows rapidly as n increases as shown in Table 22

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Table 22 n 1 2 3 4 5 6 7 8 9 10 n! 1 2 6 24 120 720 5,040 40,320 362,880 3,628,800

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Permutations with Some Objects Alike

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Sometimes not all of the objects to be permuted are distinct For example, suppose we have 3, A s, 4 B s, and 5 C s to be permuted, or 12 objects all together There are not 12! permutations, since the A s are not distinguishable from each other, nor are the B s, nor are the C s Suppose we let G be the number of distinct permutations and that we have a list of these permutations Now number the A s from 1 to 3These can be permuted in 3! ways; so, if we permute the A s in each item in our list, the list now has 3!G items Now label the B s from 1 to 4 and permute the B s in each item in the list in all 4! ways The list now has 4!3!G items Finally, number the 5 C s and permute these for each item in the list The list now contains 5!4!3!G items But now each of the items is distinct,

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so the list has 12! items We see that 5!4!3!G = 12!, so G = and this is considerably less than 12! = 479, 001, 600

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12! 5!4!3!

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or G = 27, 720

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Permuting Only Some of the Objects

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Now suppose that we have n distinct objects and we wish to permute r of them, where r n We now have r boxes to ll This can be done in n (n 1) (n 2) [n (r 1)] = n (n 1) (n 2) (n r + 1) ways If r < n, this expression is not a factorial, but can be expressed in terms of factorials by multiplying and dividing by (n r)! We see that n (n 1) (n 2) (n r + 1) = = n (n 1) (n 2) (n r + 1) (n r)! (n r)! n! (n r)!

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We will have little use for this formula We derived it so that we can count the number of samples that can be chosen from a population, which we do subsequently For the formula to work for any value of r, we de ne 0! = 1 We remark now that the 20 applicants to the executive faced with choosing a new assistant could appear in 20! = 24, 329, 020, 081, 766, 400, 000 different orders Selecting the best of the group by making a random choice means that the best applicant has a 1/20 = 005 chance of being selected, a fairly low probability So the executive must create a better procedure The executive can, as we will see, choose the best candidate with a probability approaching 1/3, but that is something we will discuss much later There are 52! distinct arrangements of a deck of cards This number is of the order 8 1067 It is surprising to nd, if we could produce 10, 000 distinct permutations of these per second, that it would take about 2 1056 years to enumerate all of these We usually associate impossible events with in nite sets, but this is an example of a nite set for which this event is impossible For example, suppose we have four objects (a, b, c, and d again) and that we wish to permute only two of these We have four choices for the leftmost position and three choices for the second position, giving 4 3 = 12 permutations Applying the formula we have n = 4 and r = 2, so

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4! 4! 4 3 2! = = = 4 3 = 12 (4 2)! 2! 2!

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giving the correct result Permutations are often the basis for a sample space in a probability problem Here are two examples

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