From Carol Ash's, "Probability Tutoring Book", pg. 19, prob. 1-3.4.
If a 12 symbol string is formed from the 10 digits and 26 letters, repetition not allowed, what is the prob that it contains 3 even digits?
I assumed string implies order matters, so I counted of $P(36,12)=\frac{36!}{(36-12)!}$ total possible strings.
Then for the numerator I counted:
- Pick 3 spots for the 3 even digits
- Fill the first, second and third even digit slots
- From the remaining 31 non-even number characters pick something for each of the remaining 9 slots.
So I got $\frac{\binom{12}{3}\times P(5,3)\times P(31, 9)}{P(36,12)}$ where $P(n,k)=\frac{n!}{(n-k)!}$.
However Ash says:
For the total, pick a committee of 12 symbols from the 36. For the fav, pick a subcommittee of 3 evens and a subcommittee of 9 others. $\frac{\binom{5}{3}\binom{31}{9}}{\binom{36}{12}}$
My question is:
- Did I misunderstand the question and order doesn't matter?
- Order does matter and I'm just wrong
- These actually work out the same ? ( if so why? )