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Problem 46. While the formula you proved in Problems 35 and 39.d is very useful, it doesn't give us a sense of how big the binomial coefficients are. We can get a very rough idea, for example, of the size of (²") by recognizing that we can write (2n)n/n! as 2n n 2n- n-1 1 n+1, 1 and each quotient is at least 2, so the product is at least 2n. If this were an accurate estimate, it would mean the fraction of n-element subsets of a 2n-element set would be about 2"/2²M = 1/2", which becomes very small as n becomes large. However it is pretty clear the approximation will not be a very good one, because some of the terms in that product are much larger than 2. In fact, if 2n ) were the same for every k, then each would be the fraction 1 of 22M. This is much 2n+1 larger than the fraction 1/2 But our intuition suggets that (2" ) is much larger than (4) and is likely larger than n-1 2n ) so we can be sure our approximation is a bad one. For estimates like this, James Stirling developed a formula to approximate n! when n is large, namely n! is about (V2nn)nnjen. In fact the ratio of n! to this expression approaches 1 as n becomes infinite. We write this as 11-V2TH a We read this notation as n! is asymptotic to 2nn Use Stirling's for- mula to show that the fraction of subsets of size n in an 2n-element set is approximately This is a much bigger fraction than 1/2

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