Permutations with Repetitions Multiplicative Identity: 1a = a

Permutations with Repetitions 79

5.5 Permutations with Repetitions

We now consider permutations with repeated objects. 518 Example In how many ways may the letters of the word MASSACHU SETT S be permuted? Solution: We put subscripts on the repeats forming MA 1 S 1 S 2 A 2 CHU S 3 ET 1 T 2 S 4 . There are now 13 distinguishable objects, which can be permuted in 13 different ways by Theorem 507. For each of these 13 permutations, A 1 A 2 can be permuted in 2 ways, S 1 S 2 S 3 S 4 can be permuted in 4 ways, and T 1 T 2 can be permuted in 2 ways. Thus the over count 13 is corrected by the total actual count 13 242 = 64864800 . A reasoning analogous to the one of example 518, we may prove 519 Theorem Let there be k types of objects: n 1 of type 1; n 2 of type 2; etc. Then the number of ways in which these n 1 + n 2 + ··· + n k objects can be rearranged is n 1 + n 2 + ··· + n k n 1 n 2 ···n k . 520 Example In how many ways may we permute the letters of the word MASSACHU SETT S in such a way that MASS is always together, in this order? Solution: The particle MASS can be considered as one block and the 9 letters A , C, H, U, S, E, T, T, S. In A, C, H, U, S, E, T, T , S there are four S’s and two T ’s and so the total number of permutations sought is 10 22 = 907200 . 521 Example In how many ways may we write the number 9 as the sum of three positive integer summands? Here order counts, so, for example, 1 + 7 + 1 is to be regarded different from 7 + 1 + 1. Solution: We first look for answers with a + b + c = 9 , 1 ≤ a ≤ b ≤ c ≤ 7 80 Chapter 5 and we find the permutations of each triplet. We have a , b, c Number of permutations 1 , 1, 7 3 2 = 3 1 , 2, 6 3 = 6 1 , 3, 5 3 = 6 1 , 4, 4 3 2 = 3 2 , 2, 5 3 2 = 3 2 , 3, 4 3 = 6 3 , 3, 3 3 3 = 1 Thus the number desired is 3 + 6 + 6 + 3 + 3 + 6 + 1 = 28 . 522 Example In how many ways can the letters of the word MURMUR be arranged without letting two letters which are alike come together? Solution: If we started with, say , MU then the R could be arranged as follows: M U R R , M U R R , M U R R . In the first case there are 2 = 2 of putting the remaining M and U, in the second there are 2 = 2 and in the third there is only 1. Thus starting the word with MU gives 2 + 2 + 1 = 5 possible arrangements. In the general case, we can choose the first letter of the word in 3 ways, and the second in 2 ways. Thus the number of ways sought is 3 · 2 · 5 = 30. 523 Example In how many ways can the letters of the word AFFECTION be arranged, keeping the vowels in their natural order and not letting the two F’s come together? Solution: There are 9 2 ways of permuting the letters of AFFECTION. The 4 vowels can be permuted in 4 ways, and in only one of these will they be in their natural order. Thus there are 9 24 ways of permuting the letters of AFFECTION in which their vowels keep their natural order. Now, put the 7 letters of AFFECTION which are not the two F’s. This creates 8 spaces in between them where we put the two F’s. This means that there are 8 · 7 permutations of AFFECTION that keep the two F’s together. Hence there are 8 · 7 4 permutations of AFFECTION where the vowels occur in their natural order. In conclusion, the number of permutations sought is 9 24 − 8 · 7 4 = 8 4  9 2 − 1 ‹ = 8 · 7 · 6 · 5 · 4 4 · 7 2 = 5880 Homework 81 524 Example How many arrangements of five letters can be made of the letters of the word PALLMALL? Solution: We consider the following cases: ➊ there are four L’s and a different letter. The different letter can be chosen in 3 ways, so there are 3 · 5 4 = 15 permutations in this case. ➋ there are three L’s and two A’s. There are 5 32 = 10 permutations in this case. ➌ there are three L’s and two different letters. The different letters can be chosen in 3 ways either P and A; or P and M; or A and M, so there are 3 · 5 3 = 60 permutations in this case. ➍ there are two L’s, two A’s and a different letter from these two. The different letter can be chosen in 2 ways. There are 2 · 5 22 = 60 permutations in this case. ➎ there are two L’s and three different letters. The different letters can be chosen in 1 way. There are 1 · 5 2 = 60 permutations in this case. ➏ there is one L. This forces having two A’s and two other different letters. The different letters can be chosen in 1 way. There are 1 · 5 2 = 60 permutations in this case. The total number of permutations is thus seen to be 15 + 10 + 60 + 60 + 60 + 60 = 265 . Homework 525 Problem In how many ways may one permute the letters of the word MEPHISTOPHELES? 526 Problem How many arrangements of four letters can be made out of the letters of KAFFEEKANNE without letting the three E’s come together? 527 Problem How many numbers can be formed with the digits 1 , 2, 3, 4, 3, 2, 1 so that the odd digits occupy the odd places? 528 Problem In this problem you will determine how many different signals, each consisting of 10 flags hung in a line, can be made from a set of 4 white flags, 3 red flags, 2 blue flags, and 1 orange flag, if flags of the same colour are identical. ➊ How many are there if there are no constraints on the order? ➋ How many are there if the orange flag must always be first? ➌ How many are there if there must be a white flag at the beginning and another white flag at the end? 529 Problem In how many ways may we write the number 10 as the sum of three positive integer summands? Here order counts, so, for example, 1 + 8 + 1 is to be regarded different from 8 + 1 + 1. 530 Problem Three distinguishable dice are thrown. In how many ways can they land and give a sum of 9? 531 Problem In how many ways can 15 different recruits be divided into three equal groups? In how many ways can they be drafted into three different regiments?

5.6 Combinations without Repetitions