3

How many arrangements are there of the word $STATISTICS$ such that there are:

  1. No consecutive $S$s
  2. Vowels in alphabetical order
  3. $T$'s are consecutive

I've noticed similar questions, but none that define similar rules. In my notes I have the solution as ${5 \choose 2} 2! {6 \choose 3}$, but I think that is wrong.

The first observation we can make is that $AII$ is the only ordering such that rule (2) is not violated. We can also treat $TTT$ as one character, following rule (3).

Ignoring the $S$s for a moment, we have 4 spaces between $AII$ to fit 2 characters, $TTT$ and $C$. This yields ${4 \choose 2}$ possible options.

We multiply by $2!$ because $TTT$ and $C$ can be switched in place.

So now we have some permutation following the form of: $TTTAIIC$ yielding 6 spots to fill with the 3 $S$s, so ${6 \choose 3}$ possible options.

So I think the solution is:

${4 \choose 2} 2! {6 \choose 3}$, not ${5 \choose 2}2!{6 \choose 3}$.

Is my reasoning sound and my original notes wrong? Thanks for your help!

Community
  • 1
James Taylor
  • 588

2 Answers2

2

Your reasoning is incorrect. It is not true that we have four spaces in AII to fit two characters. You seem to be assuming that C and TTT cannot be consecutive, which is not the case. Notice that ACTTTII is a valid arrangement of the letters before the S's are inserted.

Here is another method of thinking about the problem. If we set aside the three S's for the moment, we have five objects to arrange. They are an A, a C, two I's, and a block consisting of three T's. We can choose two of the five positions for the I's in $\binom{5}{2}$ ways, then arrange the remaining three objects in $3!$ ways. However, by symmetry, in only one third of these arrangements does the A precede both I's. Hence, the number of permissible ways of arranging these objects is $$\frac{1}{3}\binom{5}{2}3! = \binom{5}{2}2!$$ To ensure no two S's are consecutive, we must choose three of the six available spaces (four between successive objects in a permissible arrangement of A, C, I, I, and TTT and the two ends of the row) in which to insert an S, which can be done in $\binom{6}{3}$ ways. Hence, the number of arrangements that satisfy the requirements is, in fact, $$\binom{5}{2}2!\binom{6}{3}$$

N. F. Taussig
  • 76,571
  • 1
    Ah! Thank you for pointing out my invalid assumption. It also made sense for me to think about the problem as ${4 \choose 1}{5 \choose 1}{6 \choose 3}$ which is also the same as ${5 \choose 2}2!{6 \choose 3}$. – James Taylor Dec 11 '16 at 01:54
1

You can regard the triple $T$ as a single letter $\fbox{$TTT$}$ for the purpose of arrangement, as you say. You can approach the problem in the following order:

  • arrange $C$ and $\fbox{$TTT$}$ - $2! = 2$ options.
  • select the vowels positions (fixed ordering) - $\binom{5}{3} = 10$ options
  • choose which $3$ gaps will get a single $S$ - $\binom{6}{3} = 20$ options

giving $400$ options when multiplied up. The difference from your argument is that you are implicitly requiring a vowel between $C$ and $\fbox{$TTT$}$, but that is not a constraint.

Joffan
  • 39,627