The two ways of placing two rooks

Rook Polynomials

Combinatorics is the study of countable discrete objects, where these objects may not necessarily be concrete, but may be abstract mathematical entities, such as, paths in network. It involves many areas of modern mathematical research. Within a statistics and probability context, it involves arrangements (permutations) and selections (combinations) of like and unlike objects. However, the topic can involve more sophisticated approaches.

How many ways can eight non-taking rooks be placed on a chess board? This well-known problem is based on the positioning of the eight rooks so that no two rooks are in the same row or column. The answer is 8!

This problem can be generalised to how many ways are there of placing rooks on a by square board, where .

Taking the simplest cases:

If , then there is one way of placing 0 rooks and one way of placing 1 rook, that is, we can write

and .

If , then there is one way of placing 0 rooks, four ways of placing 1 rook and two ways of

placing 2 rooks, that is, , and .

Worked example

If , then there is one way of placing 0 rooks, nine ways of placing 1 rook, eighteen ways of

placing 2 rooks and six was of placing 3 rooks, that is, , , and .

(i) Without drawing diagrams and listing the possible arrangements, explain why

(ii) Similarly, explain why

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Solution

(i) There are 9 ways of choosing where the first rook is positioned, but the second can only be placed in one of four squares because the row and column of the first rook’s position are not allowed. The third rook can only placed in the square which is left when the rows and columns of

the first and second are excluded. So the number of arrangements is .

But, the 3 rooks are identical, or like, and so we need to reduce this number by dividing by the

number of arrangements of 3 objects, that is , and so we arrive at as the number of allowable positions for 3 rooks.

(ii) Using a similar argument, we have arrangements of 2 rooks, but the two rooks are

identical so we arrive at as the number of allowable positions for 2 rooks.

This information can be summarised by a generating function which is called the rook polynomial.

So, using the notation , we have:

and , where is the dummy variable and its indices and coefficients carry the information.

Exercise 1

1.1 Show that

1.2 Show that

1.3 Find

1.4 Find an expression for the coefficient of in

Non-square boards

So far we have look at square boards, but generally we can use the same notation for any shape of

board, that is , where is the maximum number of non-taking rooks that can be placed on the board B.

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Considering board D, there are clearly seven ways of placing 1 rook. It is also relatively easy to see that there are two ways of placing 3 rooks. (One of the rooks needs to be in the single square attached to the 3 by 2 rectangle.)

When trying to find how many ways there are of placing 2 rooks, it is helpful to consider two mutually exclusive ways: either with the 2 rooks placed solely in the 3 by 2 rectangle, or with one of the rooks placed in the single attached square, that is, six ways + four ways, making ten in total.

So .

Exercise 2

2.1 For board A, find

2.2 For board B, find

2.3 For board C, find

Recurrence rules

So far the rook polynomials have summarized information about the board and it was found by counting various arrangements. However, there are two recurrence rules that can be used to form the rook polynomials of more complex composite boards.

Rule 1: If board B comprises two non-interfering sub-boards B1 and B2,

then

For example, . This can be

expanded to give .

So there are sixteen ways of placing 2 rooks. With a little thought, one can see that the 16 arises through four combinations:

1 rook placed in the 1 by 1 sub-board and 1 rook in the 2 by 2 sub-board; 1 rook placed in the 2 by 2 sub-board and 1 rook in 1 by 2 sub-board;

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1 rook placed in the 1 by 2 sub-board and 1 rook in the 1 by 1 sub-board; or 2 rooks in the 2 by 2 sub-board.

This corresponds to which, in turn, is how the coefficient of arises when the three factors are multiplied.

Worked example

(i) Explain how the coefficient of in arises in three combinations and relate these to the three non-interfering sub-boards.

(ii) Explain how the coefficient of in arises in one combination and relate this to the three non-interfering sub-boards.

Solution

(i) There are three combinations: 1 rook in the 1 by 1 and 2 in the 2 by 2; 1 in the 1 by 1, 1 in the 2 by 2 and 1 in the 1 by 2; or 2 in the 2 by 2 and 1 in the 1 by 2. This corresponds to (1)(2)+(1)(4)(2)+(2)(2)=14

(ii) There is only one combination: 1 rook in the 1 by 1, 2 in the 2 by 2 and 1 in the 1 by 2. This corresponds to (1)(2)(2)=4

Rule 2: If a particular square in board B is chosen, and sub-boards B1 and B2 are formed by

(1) deleting the row and column that contain the particular square, and

(2) deleting just the square alone,

then

For example, consider the bottom left square in board D and the chosen square. Boards D1 and D2 can be formed as shown below.

By inspection it can be shown that and .

So .

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This can seen by looking at the problem with (i) the first rook placed in the chosen square and the remaining 2 placed in D1, or (ii) the chosen square not having a rook and all the rooks placed in D2.

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Exercise3

3.1 Find the rook polynomial for board F.

3.2 Find the rook polynomial for board G

3.3 By using and rule 2 in reverse, confirm your answer for found in the previous question.

3.4 Find the rook polynomial for board H.

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3.5 Find the rook polynomial for board I.

Forbidden positions

The two recurrence rules can be very helpful, but actual situations arise more often by not having a particular irregular composite board, but by having a square board with forbidden positions. The simplest of these arise in considering derangements.

Consider, how many ways can four letters be delivered to 4 houses without any arriving at its correct house? The answer is given by the formula for number of derangements of four

objects

This is usually derived by using the Inclusion-Exclusion Principle, but it is equivalent to placing 4 rooks in the board opposite. The forbidden positions are those that are crossed out. We can work out the rook polynomial for these forbidden positions by using the first recurrence rule, that is,

, that is, .

Somewhat unbelievable, the number of ways for the allowed (or non-forbidden) positions is found

by replacing by . That is,

Not only does this work for the derangement board above, it also works more generally. So, in general, we have what is called the Rook Inclusion Exclusion Theorem (Rook PIE), that is,

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If , then the number of ways of arranging rooks in the allowed

positions is .

Why does this procedure work? As its name implies, it is proved by using the Inclusion-Exclusion

Principle. The comes from the alternate subtracting or adding of the size of multiple intersections of sets of arrangements of one rook in one of the forbidden squares and the other

placed in the allowable positions. In other words, we have

The power of this procedure is that it works when the forbidden positions are more irregular than just the leading diagonal.

Worked example

In a given company, worker A cannot do jobs a and b, worker B cannot do job b, and workers C and D can do neither c nor d. In how many ways can the workers be assigned to the jobs?

Solution

The rook polynomial for the board forbidden positions (bfp) is

, which is

So the number of different assignments is .

[Note: clearly, in this simple example, it would have been easier to work out the four ways by inspection, that is, C and D are restricted to jobs a and b and they can be assigned two different ways. Similarly, A and B have to do jobs c and d, and they can be assigned to them two different ways. So now by the multiplication principle, the answer is 4.]

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Exercise 4

4.1 For the board opposite

(i) Find the rook polynomial for the board of forbidden positions

(ii) Find the number of ways of placing 5 rooks in the allowed positions.

4.2 Alan, Bob, Charlie, Rose, Susan and Tanya are enrolled for Salsa classes. The organiser asks each girl confidentially to share their dancing partner preferences. Rose prefers not to partner Charlie and Tanya similarly won’t partner Alan. Susan is happy to partner any of the boys.

(i) Find the rook polynomial for the board of forbidden positions.

(ii) Find how many ways the dancing partners can be arranged.

4.3 Anne, Betty, Chris and David have been elected as a school’s charity committee. Chris and David refuse to be chairperson. Anne doesn’t want to be treasurer and Betty doesn’t want to be the secretary or the fund raiser. Chris also doesn’t want to be secretary. In how many ways can the four positions on the committee be filled?

4.4 New recruits a, b, c, d and e to a company are to be mentored by established employees R, S, T, U and V. The Human Resources manager reviews the curriculum vitae of the recruits and decides that R cannot mentor b or c. T can only mentor d or e. S can’t mentor b, U can’t mentor e and V can’t mentor d. How many ways can the recruits be matched with mentors?

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Answers to exercises

Exercise 1

1.1 There are ways of placing one rook. In the case of 2 rooks, the first can be positioned

ways and the second ways , but they are identical so the is divided by . So

. In the case of 3 rooks, we have divided by . So

. In the case of 4 rooks, we have .

1.2 By similar arguments as in 1.1, we have , ,

, and

1.3

1.4

Exercise 2

2.1

2.2

2.3

Exercise 3

3.1

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3.2

3.3

3.4

3.5

Exercise 4

4.1 (i) (ii) 16 ways

4.2 (i) (ii) 3 ways

4.3 4 ways

4.4 14 ways

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