Probability and Statistical Inference. Robert Bartoszynski
alt="images"/> there are 10 factors, and each term will contain one component from each set of parentheses. Thus, choosing
An approximation to
Theorem 3.4.3 (Stirling's Formula) We have
where the sign
We shall not give the proof here, but interested readers can find it in more advanced texts, for example, in Chow and Teicher (1997).
Example 3.15
A group of
Solution
Clearly, the number of ways a group can be divided into two equal parts is
(3.30)
which, based on (3.29), can be approximated by
For example, in the case of 16 boys and 16 girls
Problems
1 3.4.1 Show that if then(i) Use the definition of binomial coefficients as ratios of the factorials. (ii) Use directly the interpretation of the binomial coefficients as the number of subsets of a given size. (iii) How many ways can one choose an ‐element subset from a ‐element subset from a ‐element subset from a ‐element subset from a element set? (where ).
2 3.4.2 Find the coefficient of the term in the expansion of .
3 3.4.3 Use the argument analogous to that in Theorem 3.3.2 to show that if , and , then
4 3.4.4 Use Stirling's formula to approximate the number of ways: (i) A set of size can be partitioned into two equal parts. (ii) A set of size can be partitioned into three equal parts.
5 3.4.5 Approximate the probability that every state will be represented in the committee in Problem 3.3.1 (i).
Notes
1 1 The integer part of , [], is the largest integer not exceeding ( etc.).
2 2 This property, called exchangeability of events, will be discussed in more detail in Section 4.6.
3 3 The discarded cards are not mixed with the deck. Assume that the player receives the replacement of the discarded cards from the unused remainder of the deck.
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