Combinatorial Methods

247 Example The strictly positive integers are written in succession

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, . . .

Which digit occupies the 3000-th position?

Solution: Upon using

a total of 9 + 180 + 2700 = 2889 digits have been used, so the 3000-th digit must belong to a 4-digit integer. There remains to use

3000 −2889 = 111 digits, and 111 = 4 ·27 + 3, so the 3000-th digit is the third digit of the 28-th 4-digit integer, that is, the third digit of 4027, namely 2.

6.2 Combinatorial Methods

Most counting problems we will be dealing with can be classifi ed into one of four categories. We explain such categories by means of an example.

248 Example Consider the set {a, b, c, d}. Suppose we “select” two letters from these four. Depending on our interpretation, we may obtain the following answers.

Permutations with repetitions. The order of listing the letters is important, and repetition is allowed. In this case there are 4 ·4 = 16 possible selections:

Permutations without repetitions. The order of listing the letters is important, and repetition is not allowed. In this case there are 4 ·3 = 12 possible selections:

Combinations with repetitions. The order of listing the letters is not important, and repetition is allowed. In this case there are

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Combinations without repetitions. The order of listing the letters is not important, and repetition is not allowed. In this case there

are 4 ·3 = 6 possible selections: 2

We will now consider some examples of each situation.

6.2.1 Permutations without Repetitions

249 Definition We define the symbol ! (factorial), as follows: 0! = 1, and for integer n ≥ 1,

n! = 1 ·2 ·3 ···n.

n! is read n factorial.

250 Example We have

251 Example Write a code fragment to compute n!.

Solution: The following is an iterative way of solving this problem.

Algorithm 6.2.1: FACTORIAL(n)

comment: returns n!

m ← 1

:><8while n > 1

m← n m

n← n −1

return (m)

252 Definition Let x1, x2, . . . , xn be n distinct objects. A permutation of these objects is simply a rearrangement of them.

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253 Example There are 24 permutations of the letters in MAT H, namely

MAT H MAHT MTAH MT HA MHTA MHAT

AMT H AMHT AT MH AT HM AHT M AHMT

TAMH TAHM T MAH T MHA T HMA T HAM

HAT M HAMT HTAM HT MA HMTA HMAT

254 Theorem Let x1, x2, . . . , xn be n distinct objects. Then there are n! permutations of them.

Proof: The first position can be chosen in n ways, the second object in n −1 ways, the third in n −2, etc. This gives

n(n −1)(n −2) ··· 2 ·1 = n!.

255 Example Write a code fragment that prints all n! of the set {1, 2, . . . , n}.

Solution: The following programme prints them in lexicographical order. We use examples 13 and 23.

Algorithm 6.2.2: PERMUTATIONS(n)

k ← n −1

§while X [k] > X [k −1]

k ← k −1

t ← k + 1

§while ((t < n) and (X [t + 1] > X [k]))

t ← t + 1

comment: now X [k + 1] > . . . > X [t] > X [k] > X [t + 1] > . . . > X [n]

Swap(X [k], X [t])

comment: now X [k + 1] > . . . > X [n]

ReverseArray(X [k + 1], . . . , X [n])

256 Example A bookshelf contains 5 German books, 7 Spanish books and 8 French books. Each book is different from one another.

Solution:

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257 Example Determine how many 3-digit integers written in decimal notation do not have a 0 in their decimal expansion. Also, find the sum of all these 3-digit numbers.

Solution: There are 9 ·9 ·9 = 729 3-digit integers not possessing a 0 in their decimal expansion. If 100x + 10y + z is such an integer, then given for every fixed choice of a variable, there are 9 ·9 = 81 choices of the other two variables. Hence the required sum is

81(1 + 2 + ·+ 9)100 + 81(1 + 2 + ·+ 9)10 + 81(1 + 2 + ·+ 9)1 = 404595.

258 Example Determine how many 3-digit integers written in decimal notation possess at least one 0 in their decimal expansion. What is the sum of all these integers.

Solution: Using example 257, there are 900 −729 = 171 such integers. The sum of all the three digit integers is

100 + 101 + ··· + 998 + 999.

To obtain this sum, observe that there are 900 terms, and that you obtain the same sum adding backwards as forwards:

=900(1099),

giving S = 900(1099) = 494550. The required sum is 494550 −404595 = 89955. 2

6.2.2 Permutations with Repetitions

We now consider permutations with repeated objects.

259 Example In how many ways may the letters of the word

MASSACHU SET T S

be permuted?

Solution: We put subscripts on the repeats forming

MA1S1S2A2CHU S3ET1T2S4.

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There are now 13 distinguishable objects, which can be permuted in 13! different ways by Theorem 254. For each of these 13! permutations, A1A2 can be permuted in 2! ways, S1S2S3S4 can be permuted in 4! ways, and T1T2 can be permuted in 2! ways. Thus the over count 13! is corrected by the total actual count

13!

= 64864800.

2!4!2! A reasoning analogous to the one of example 259, we may prove

260 Theorem Let there be k types of objects: n1 of type 1; n2 of type 2; etc. Then the number of ways in which these n1 + n2 + ···+ nk

objects can be rearranged is

(n1 + n2 + ···+ nk )! . n1!n2! ···nk !

261 Example In how many ways may we permute the letters of the word MASSACHU SET T 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

2!2!10! = 907200.

262 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

and we find the permutations of each triplet. We have

Thus the number desired is

3 + 6 + 6 + 3 + 3 + 6 + 1 = 28.

263 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:

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