Can you please expound and simplify the embolden phrase below?
>As another example, with n = 365 days in a year and k people, how many possible
unordered birthday lists are there? For example, for k = 3, we want to count lists
like (May 1, March 31, April 11), where all permutations are considered equivalent.
We can't do a simple adjustment for overcounting such as $\dfrac{n^k}{3!}$ since, e.g., there are
6 permutations of (May 1, March 31, April 11) but only 3 permutations of (March
31, March 31, April 11). By Bose-Einstein, the number of lists is $\dbinom{n+k-1}{k}$. **But the
ordered birthday lists are equally likely, not the unordered lists [Emphasis mine]**, so the Bose-Einstein
value should not be used in calculating birthday probabilities. $\square$
Blitzstein. *Introduction to Probability* (2019 2 ed). p 20.
DNB
·
2021-07-09T06:29:58Z (over 3 years ago)