By Peter J. Cameron
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2 The q-binomial theorem The q-binomial coefficients satisfy an analogue of the recurrence relation for binomial coefficients. 3 n 0 = q n n n k = 1, q = q n−1 k−1 n−1 k + qk q for 0 < k < n. q Proof This comes straight from the definition. Suppose that 0 < k < n. Then n k − q n−1 k−1 qn − 1 −1 qk − 1 = q = qk = qk n−1 k−1 qn−k − 1 qk − 1 n−1 k−1 q q n . k−1 q The array of Gaussian coefficients has the same symmetry as that of binomial coefficients. From this we can deduce another recurrence relation.
Now recall the cycle decomposition of permutations: Any permutation of a finite set can be written as the disjoint union of cycles, uniquely up to the order of the factors and the choices of starting points of the cycles. Moreover, Two permutations are equivalent if and only if the lists of cycle lengths of the two permutations (written in non-increasing order) are equal. Thus equivalence classes of permutations correspond to partitions of the integer n. This means that the enumeration theory for “unlabelled permutations” is the same as that for “unlabelled partitions”, discussed in the last section.
Show that the number of ways of selecting k objects from a set of n distinguished objects, if we allow the same object to be chosen more than once and pay n+k−1 no attention to the order in which the choices are made, is . 2. Prove that, if n is even, then n 2n ≤ ≤ 2n . n+1 n/2 Use Stirling’s formula to prove that n ∼ n/2 2n . πn/2 How accurate is this estimate for small n? 3. Use the method of the preceding exercise, together with the Central Limit Theorem, to deduce the constant in Stirling’s formula.
Notes on counting [Lecture notes] by Peter J. Cameron
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