MATH 245 CH15 - DISCRETE MATHEMATICS W W L CHEN c W W L...

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DISCRETE MATHEMATICS W W L CHEN c ± W W L Chen, 1992, 2008. This chapter is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners. Chapter 15 NUMBER OF SOLUTIONS OF A LINEAR EQUATION 15.1. Introduction Example 15.1.1. Suppose that 5 new academic positions are to be awarded to 4 departments in the university, with the restriction that no department is to be awarded more than 3 such positions, and that the Mathematics Department is to be awarded at least 1. We would like to find out in how many ways this could be achieved. If we denote the departments by M,P,C,E , where M denotes the Mathematics Department, and denote by u M ,u P ,u C ,u E the number of positions awarded to these departments respectively. Then clearly we must have u M + u P + u C + u E = 5 . (1) Furthermore, u M ∈ { 1 , 2 , 3 } and u P ,u C ,u E ∈ { 0 , 1 , 2 , 3 } . (2) We therefore need to find the number of solutions of the equation (1), subject to the restriction (2). In general, we would like to find the number of solutions of an equation of the type u 1 + ... + u k = n, where n,k N are given, and where the variables u 1 ,...,u k are to assume integer values, subject to certain given restrictions. Chapter 15 : Number of Solutions of a Linear Equation page 1 of 13
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Discrete Mathematics c ± W W L Chen, 1992, 2008 15.2. Case A – The Simplest Case Suppose that we are interested in finding the number of solutions of an equation of the type u 1 + ... + u k = n, (3) where n,k N are given, and where the variables u 1 ,...,u k ∈ { 0 , 1 , 2 , 3 ,... } . (4) PROPOSITION 15A. The number of solutions of the equation (3) , subject to the restriction (4) , is given by the binomial coefficient ± n + k - 1 n ² . Proof. Consider a row of ( n + k - 1) + signs as shown in the picture below: + + + + + + + + + + + + + + + + ... + + + + + + + + + + + + + + + + | {z } n + k - 1 Let us choose n of these + signs and change them to 1’s. Clearly there are exactly ( k - 1) + signs remaining, the same number of + signs as in the equation (3). The new situation is shown in the picture below: 1 ... 1 | {z } u 1 + ... + 1 ... 1 | {z } u k | {z } k (5) For example, the picture +1111 + +1 + 111 + 11 + ... + 1111111 + 1111 + 11 denotes the information 0 + 4 + 0 + 1 + 3 + 2 + ... + 7 + 4 + 2 (note that consecutive + signs indicate an empty block of 1’s in between, and the + sign at the left-hand end indicates an empty block of 1’s at the left-hand end; similarly a + sign at the right-hand end would indicate an empty block of 1’s at the right-hand end). It follows that our choice of the n 1’s corresponds to a solution of the equation (3), subject to the restriction (4). Conversely, any solution of the equation
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MATH 245 CH15 - DISCRETE MATHEMATICS W W L CHEN c W W L...

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