Pemutations of Permutation

In the arrangement of the word PERMUTATION , there are 5! ways to arrange the 5 vowels in the word. Only 1 out of these 5! ways are correct.

Therefore, out of the 11!/2! possible arrangement of the letters in the word PERMUTATION with repeated letter T, only 1/5! of these arrangements have the 5 vowels in alphabetical order.

Thus, the number of ways to arrange PERMUTATION, keeping the vowels in alphabetical order, though not necessarily together is 11!/2!5! = 166320.

Think of the vowels as "walls".

$(a)\quad \quad \quad A\quad \quad \quad (b)\quad \quad \quad E\quad \quad \quad (c)\quad \quad \quad I\quad \quad \quad (d)\quad \quad \quad O\quad \quad \quad (e)\quad \quad \quad U\quad \quad \quad (f)$

There are 11 letters in the word "PERMUTATION", and 5 are already arranged, as above since they have to be in a specific order. So, the remaining 6 letters, P, R, M, T, T, N, has to be partitioned, and arranged in the spaces shown above; (a), (b), (c), (d), (e).

So first, we find the total number of ways to partition the letters into each space. Why? Because the number of letters that fit in each space is different for each case. For example, the letters might be partitioned equally, or 3 letters might be in one space, or all 6 letters in one space, as shown below.

$(P)\quad \quad \quad A\quad \quad \quad (M)\quad \quad \quad (T)\quad \quad \quad E\quad \quad \quad (T)\quad \quad \quad I\quad \quad \quad (N)\quad \quad \quad O\quad \quad \quad (R)\quad \quad \quad U\\ (P)\quad \quad \quad (M)\quad \quad \quad (T)\quad \quad \quad A\quad \quad \quad (T)\quad \quad \quad E\quad \quad \quad (N)\quad \quad \quad I\quad \quad \quad (R)\quad \quad \quad O\quad \quad \quad U\\ (P)\quad \quad \quad (M)\quad \quad \quad (T)\quad \quad \quad (T)\quad \quad \quad (N)\quad \quad \quad (R)\quad \quad \quad A\quad \quad \quad E\quad \quad \quad I\quad \quad \quad O\quad \quad \quad U\\ ...$

To do this, let's put aside the letters for a second. Think that we have to partition 6 1's in each space, separated by "|". Like this:

$(1)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\\ (1)\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\\ ...$

If we assume that if 2 or more 1's are next to each other, we add them,

$(1)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\\ (2)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\quad \quad \quad |\quad \quad \quad (1)\\ ...$

Now, to find the total number of ways to partition the letters into each space from the very first diagram, think of it as arranging 6 1's and 5 "|"s, which is 11!/(6!)(5!) <We made a quick assumption from above that 1's were the consonants, and "|"'s were vowels.>

Now, we just have to multiply it by the permutation of the 6 consonants, and we're done!

$\frac { 11! }{ 6!\times 5! } \times \frac { 6! }{ 2 } =166320\quad (ways)$

There are 5 vocals and 6 consonants, the number of arrangements: ${11}\choose{6}$ or ${11}\choose{5}$

since the internal arrangement of the vocals is fixed, the internal arrangement of the consonant: $6!$

$\therefore$ number of arrangements: $\frac{11!}{5! * 2} = \boxed{166320}$