By Andrew D. Ker

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**Extra info for Discrete Mathematics [Lecture notes]**

**Example text**

This is an important example which, when specified precisely, is called observational equivalence; it does not say that P1 and P2 have the same internal working, or are equally efficient, though. When we are given an equivalence relation on A we often want to know all the elements of A which are related to some fixed a. Definition If ∼ is an equivalence relation on A, and a ∈ A, we define [a] = {a′ ∈ A | a′ ∼ a}. The sets [a], for each a, are called the equivalence classes of ∼. Sometimes, to avoid ambiguity, we might write [a]∼ if it is not clear what equivalence relation is meant.

9, the same as the number of ways of distributing 4 objects into 3 boxes: 4+3−1 = 62 = 15. 3−1 To find the largest 3-digit integer with a digit sum of 4, note that the “hundreds” digit is most significant, so make it as large as possible: the answer is 400. 5 We are dealing with 3-digit integers. For now, ignore the condition that individual digits can only be as large as 9. 9, the same as the number of ways of distributing 10 objects into 3 boxes: 10+3−1 = 12 3−1 2 = 66. Now exclude cases when a “digit” is equal to 10: there are exactly 3 such cases, because if one “digit” equals 10 then the rest must be zero.

Cancelling the fractions and noting that (n − n1 − n2 − · · · − ng )! = 0! ···ng ! By analogy with binomial coefficients, the multinomial coefficient is written n . n1 n2 · · · ng n! ···ng ! 14 If the card game bridge, all 52 cards in a pack are distributed amongst four players in a deal, with each player given 13 cards. How many different deals are there? Answer We are asked how many ways there are to split 52 cards (all different) into 4 groups of 13. By the preceding argument, the number is 52 52!