Question

Medal.


What is the probability that the 3 vowels are NOT next to each other in every arrangement of
"STATISTICS"?


Ideas: Number of ways to rearrange them is 10!/(3!3!2!)

Answer 1

What is the probability that the 3 vowels ARE next to each other in every arrangement of
"STATISTICS"?

Answer 2

Well, the 3 vowels are A I I
There are 10 positions.
The A could be in any of the 10 positions.
I'm not sure how to deal with 3 elements next to each other.

Answer 3

There are three possible arrangements of the vowels (of which there are only three): \(\dbinom3{2,1}=\dfrac{3!}{2!1!}=3\).

There are a total of eight possible ways of grouping the characters together (3 vowels, 7 consonants; 1 consonant, 3 vowels, 6 consonants; etc) - basically a star-and-bars counting problem to find how many non-negative integer solutions there are to \(a+3+b=10\), or \(a+b=7\), for \(a\) and \(b\): \(\dbinom{7+2-1}{2-1}=\dbinom81=8\). Multiplying this by \(3\) (to account for the various positions for the letter \(\mathrm A\)) gives a total of \(24\) possible arrangements for which the vowels are concatenated.

Divide this by the total number of possible arrangements and that should give you the probability you want.

Answer 4

Or rather, the complement of the probability you're looking for.

Answer 5

Thanks for your effort.

I haven't learned the star-and-bars method, or the why there are only 3 possible arrangements of the vowels...

Answer 6

There are only three vowels in "STATISTICS", with \(3!\) total permutations, but since there are two copies of "I", you have \(\dfrac{3!}{2!}=3\) possible arrangements of a three-vowel block. They are \(\mathrm {AII}\), \(\mathrm {IAI}\) and \(\mathrm {IIA}\). This count can also be obtained with the multinomial coefficient formulation in my first comment.

You can read up on the stars and bars counting method here: https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
The second theorem was what I used.

Answer 7

So then since I have 3 possible arrangements of them, I multiply it by the number of positions they can fill up?

So like since there's 10 total spaces, and we're using 3 spaces for an arrangement, then there's 8 spaces to place the 3 vowels next to each other.

So the complement is 3 * 8?

Answer 8

Cause that's 50,400 total arrangements - 24

Which seems weird to me.

Answer 9

\[\frac{ 10! }{ 2!3!3! }\] = 50,400

Answer 10

When in doubt, you can always consider a simpler example if only to make sure the counting procedure above is accurate.

Suppose you have the word \(\mathrm{GAEA}\), and you want the probability that any permutation of these letters doesn't have three consecutive vowels. There's a similar situation with getting three consecutive vowels - \(\dfrac{3!}{2!}=3\) possible arrangements, \(\mathrm{AAE},\mathrm{AEA},\mathrm{EAA}\).

The stars and bars counting problem is finding the number of non-negative solutions to \(a+3+b=4\iff a+b=1\), which is \(\dbinom{1+2-1}{2-1}=2\). In other words, you can have \(\mathrm{GV}\) or \(\mathrm {VG}\), where \(\mathrm V\) is any one of the vowel concatenations. So there are \(3\times2=6\) total ways to get three consecutive vowels.

This is a small list of possibilities, so we can write them out:
\[\begin{matrix}
\mathrm{GAAE}&\mathrm{GAEA}&\mathrm{GEAA}\\
\mathrm{AAEG}&\mathrm{AEAG}&\mathrm{EAAG}
\end{matrix}\]Meanwhile, there are \(\dbinom4{1,2,1}=\dfrac{4!}{1!2!1!}=12\) possible permutations overall (not counting repetitions), and these are
\[\begin{matrix}
\mathrm{GAAE}&\mathrm{GAEA}&\mathrm{GEAA}\\
\mathrm{AGAE}&\mathrm{AGEA}&\mathrm{AEGA}&\mathrm{AEAG}&\mathrm{AAGE}&\mathrm{AAEG}\\
\mathrm{EGAA}&\mathrm{EAGA}&\mathrm{EAAG}
\end{matrix}\]So there's a \(\dfrac{1}{2}\) probability of getting three consecutive vowels and an equal probability of that not happening. As you add more consonants to the original word, the latter probability will inevitably increase while the probability of getting consecutive vowels will diminish rapidly.