Permutations and Combinations
2.3 Aside: Making Sense of 0!
Lesson materials located below the video overview.
| Consider the “word” CHEESIESTESSNESS, the quality of being the cheesiest of cheeses. Do you see that there are\(\dfrac{16!}{5!6!}\) ways to arrange its letters? |
Exercise 13: Actually evaluate this number.
This may seem strange, but it is actually better to write this answer as:
\(\dfrac{16!}{1!1!5!6!1!1!1!}\)
\(1!\) for the one letter C
\(1!\) for the one letter H
\(5!\) for the five letters E
\(6!\) for the six letters S
\(1!\) for the one letter I
\(1!\) for the one letter T
\(1!\) for the one letter N
This offers a self check: The numbers appearing on the bottom should sum to the number appearing on the top.
Question: Back to \(1!\). Does the definition of factorial actually make sense for \(1!\)? In this context, we want \(1!\) to equal one so that the value of the denominator is still \(5!6!\). So maybe the mathematics here is suggesting that we should set \(1!=1\).
| Exercise 14: In how many ways can you rearrange the letters of your full name? |
Carrying on with CHEESIESTESSNESS …
We’ve written the formula for the number of ways to rearrange its letters as:
\(\dfrac{16!}{1!1!5!6!1!1!1!}\)
with one term in the denominator for the count of each letter in the word.
Hmmm. As there are no Ps in this word, should we a actually write this formula as: \(\dfrac{16!}{1!1!5!6!1!1!1!0!}\) ?
There are also no Qs and no Ms: \(\dfrac{16!}{1!1!5!6!1!1!1!0!0!0!}\) .
The quantity \(0!\) doesn’t actually make sense, but what value might we as a society assign to it so that the above formulas still make sense and are correct?
| Exercise 15: Evaluate the following expressions:a) \(\dfrac{800!}{799!}\) b) \(\dfrac{15!}{13!2!}\) c) \(\dfrac{87!}{89!}\) d) \(\dfrac{0!}{0!}\) |
All the answers appear in the COMPANION GUIDE to this Permutations and Combinations course.
| Exercise 16: Make the following expressions look significantly simpler:a) \(\dfrac{N!}{N!}\) b) \(\dfrac{N!}{(N-1)!}\) c) \(\dfrac{n!}{(n-2)!}\) d) \(\dfrac{1}{k+1} \cdot \dfrac{(k+2)!}{k!}\) e) \(\dfrac{n!(n-2)!}{\left((n-1)!\right)^{2}}\) |
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