# Possible arrangements of the letters of the word "Polyunsaturated" in which the vowel order is preserved

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.

• 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?
– Anar
Apr 30 at 8:10
• Apr 30 at 8:24
• Surely you must have tried something ? Pl. edit it in. Apr 30 at 8:33

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, $$\binom{15}{6}\times \frac{9!}{2!}=5005 \times 181,440=908,107,200$$

• Nice use of probability (+1) Apr 30 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 $$7.8.9.10.11.12.13.14.15$$ 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.

• The permutations are not restricted to one vowel per gap. Apr 30 at 11:33
• $+1$ for classical approach :) Apr 30 at 13:35