COMBINATORIAL COUNTING

Transcription

1 COMBINATORIAL COUNTING Our main reference is [1, Section 3] 1 Basic counting: functions and subsets Theorem 11 (Arbitrary mapping Let N be an n-element set (it may also be empty and let M be an m-element set, m 1 Then the number of all possible mappings f : N M is m n Proof Because for each element a N, f can map a to m possible values The result then follows by the product rule Theorem 12 (The total number of subsets Any n-element set X has exactly 2 n subsets (n 0 Proof This result can be proved by induction on n We can also prove it using the previous theorem with the observation that any subset of X can be assigned to a unique function f mapping from X to {0, 1} We can also prove a bit more Theorem 13 Let n 1 Each n-element set has exactly 2 n 1 subsets of an odd size and exactly 2 n 1 subsets of an even size Proof It suffices to wor with subsets of odd sizes Let us fix an element a X Any subset A X\{a} can be completed to a subset A X with an odd number of elements, by the following rule: if A is odd, we put A = A, and for A even, we put A = A {a} It is easy to chec that this defines a bijection between the system of all subsets of X\{a} and the system of all odd-size subsets of X Therefore, the number of subsets of X of odd cardinality is 2 n 1 Unlie in our first theorem where f is arbitrary, now we count the number of injective functions Theorem 14 (One-to-one mapping For given numbers n, m 0, there exist exactly n 1 m(m 1 (m n + 1 = (m i one-to-one mappings of a given n-element set to a given m-element set 1 i=0

2 2 COMBINATORIAL COUNTING Proof Let x 1,, x n be the elements of an n-element set X, and y 1,, y m be the elements of an m-element set Y Let f be a one-to-one map from X to Y Then f(x 1 can tae m different values After choosing f(x 1, there are only m 1 values that f(x 2 can tae, etc The result then follows again by the product rule 2 Binomial coefficients The binomial coefficient ( n is a function of n, given by ( n = n!!(n! = 1 i=0 (n i! Definition 21 Let X be a set and let be a nonnegative integer By the symbol ( X we denote the set of all -element subsets of the set X Our first theorem is the following Theorem 22 For any finite set X, the number of all -element subsets equals ( X Proof Put n = X We will count all ordered -tuples of elements of X (without repetitions of elements in two ways On the one hand, we now that the number of the ordered - tuples is n(n 1 (n + 1 by our last theorem of the previous section (on counting injective maps! On the other hand, from one -element subset M ( X, we can create! distinct ordered -tuples, and each ordered -tuple is obtained from exactly one -element subset M in this way Hence ( X n(n 1 (n + 1 =! We introduce some basic properties of binomial coefficients Theorem 23 For any n, (symmetric (recursion-pascal ( ( n n = n ( n ( n 1 = ( n

3 COMBINATORIAL COUNTING 3 Proof Either by formulas, or by conceptual In our next result, we give a non-trivial application of the theorem above Question 24 How many ways are there to write a nonnegative integer m as a sum of r nonnegative integer addends, where the order of addends is important? For example, for n = 3 and r = 2, we have the possibilities 3 = 0 + 3, 3 = 1 + 2, 3 = 2 + 1, and 3 = In other words, we want to find out how many ordered r-tuples (i 1, i 2,, i r of nonnegative integers there are satisfying the equation i 1 + i i r = m (1 Answer: ( m+r 1 r 1 Proof We can imagine that each of the variables i 1, i 2,, i r corresponds to one of r boxes We have m indistinguishable balls, and we want to distribute them into these boxes in some way (we assume that each box can hold all the m balls if needed Each possible distribution encodes one solution of (1 This can be interpreted as having m balls and r 1 walls separating the boxes remain Hence, choosing a distribution of the balls means selecting the position of the internal walls among the balls In other words, we have m+r 1 objects, balls and internal walls, arranged in a row, and we determine which positions will be occupied by balls and which ones by walls Thus we need to choose r 1 positions from m + r 1 positions, and this can be done in ( m+r 1 r 1 ways As we will see in later sections, having an analytical explanation of the binomial coefficient is quite useful Theorem 25 (Binomial theorem For any nonnegative integer n, we have (1 + x n = n =0 ( n x Note that this result easily implies our results in the first section on the total number of subsets, and on subsets of even/odd size Proof We just need to expand (1 + x n and see the the coefficient of x is exactly ( n The binomial coefficients can be generalized to multinomial coefficients: if we have objects of m inds, i indistinguishable objects of the i-th ind, where m = n, then the number of distinct arrangements of these objects in a row is given by the expression n! 1! 2! m!

4 4 COMBINATORIAL COUNTING This expression is usually written ( n 1, 2,, m We now state an analog of the binomial theorem Theorem 26 (Multinomial theorem (x 1 + x x m n = m=n, 1,, m 0 ( n 1, 2,, m x 1 1 x 2 2 xm m 3 Estimates This part is more or less analysis It is very useful to now how large binomial coefficients are We start with some notations Definition 31 Let f, g be real functions of a single variable defined on the natural numbers (most often we assume that the values attained by both f and g are nonnegative The notation f(n = O(g(n means that there exist constants n 0 and C such that for all n n 0, the inequality f(n Cg(n holds In this case we say that f is big-oh of g Fact 32 In the following, let C, a, α, β > 0 be some fixed real numbers independent of n We have n α = O(n β whenever α β, n C = O(a n for any a > 1, (ln n C = O(n α for any α > 0 The useful estimates follow Theorem 33 (For n! We have e(n/e n n! en(n/e n Theorem 34 For every 1 n we have ( n (n/ (en/ And lastly, Theorem 35 We have 2 2m 2 m ( 2m 22m m 2m

5 In particularly, ( 2m m /2 2m has order 1/ m COMBINATORIAL COUNTING 5 References [1] J Matouse and J Nesetril, Invitation to Discrete Mathematics, Second edition