How many possible arrangements can we sort the letters of the "Polyunsaturated" so that the vowel order is preserved in this word?

I think $p,l,y,n,s,t,r,t,d$ have $\frac{9!}{2!}$ possible arrangements, because "$t$" have $2$ repetition.

$o,u,a,u,a,e$ must sit in $10$ places between $p,l,y,n,s,t,r,t,d$ and left and right of that. But I don't know how count them so that preserved this order.

  • $\begingroup$ What do you exactly mean by "vowel order"? $_o_ _u_ _a_u_a_e_$ or just $ouauae$ should be in order from left to right? $\endgroup$
    – Anar
    Commented Apr 30, 2023 at 8:10
  • $\begingroup$ Please do not just ask. Include your work. $\endgroup$ Commented Apr 30, 2023 at 8:24
  • $\begingroup$ Surely you must have tried something ? Pl. edit it in. $\endgroup$ Commented Apr 30, 2023 at 8:33

2 Answers 2


For an another way using probability:

What is the probability that $a,a,u,u,e,o$ will be in the form of correct alphabetical order ? The answer is : $$\frac{1}{6!/(2!2!)}=\frac{1}{180}$$

Now, when we order the letters of "Polyunsaturated", $1/180$ of these arrangements will be in the form that vowels are in alphabetical order.

Then , $$\frac{15!}{2!2!2!}\times \frac{1}{180}=908,107,200$$

For the third approach:

The word "Polyunsaturated" has $15$ letters, so we can think that there are $15$ empty places for replacing these letters.Firstly, lets select $6$ place for vowels among these $15$ places by $C(15,6)$ ways. When we select these $6$ places, the letters can be replaced in only one way such that they are in the form of alphabetical order.After that, we now have $9$ places to place these $9$ constonants.We can order them $9!/2!$ ways in a line.Then, $$\binom{15}{6}\times \frac{9!}{2!}=5005 \times 181,440=908,107,200$$

  • 2
    $\begingroup$ Nice use of probability (+1) $\endgroup$ Commented Apr 30, 2023 at 14:11

Corrected, thanks to @Daniel Mathias

  • Place the 6 vowels together in a line, in the order in which they appear in the word

  • Consider the consonants as balls and insert them one by one in $$ ways. (The first ball can be placed in $7$ ways, and each ball placed creates an extra space for the next one. )

  • Divide by $2$ to take care of the repeated consonant.

  • $\begingroup$ The permutations are not restricted to one vowel per gap. $\endgroup$ Commented Apr 30, 2023 at 11:33
  • 3
    $\begingroup$ $+1$ for classical approach :) $\endgroup$ Commented Apr 30, 2023 at 13:35

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