# How many ways to arrange INSTRUCTOR

How many ways to arrange INSTRUCTOR, in which there are exactly two consonants between successive pairs of vowels.

Before any full blown solution could I get a hint towards how to consider the constraint. I am running my head in circles tryimg to figure a way to set up subsets

• Hint: Use the star and bars method. – K. Rmth Dec 7 '15 at 4:53
• Hint : when you count permutation of "egg" its not different between first "g" and second "g" then $\frac{3! }{2! }$ (2! for g's) – Amir Dec 7 '15 at 4:56
• i haven't covered stars and bars yet, still working with the basic principles and permutation and choosing – dc3rd Dec 7 '15 at 4:56
• If I'm reading this question correctly, then it boils down to I _ _ U _ _ O, where the positions of the vowels can change and the underscores can be replaced with a chosen consonant. – CaptainObvious Dec 7 '15 at 5:00
• @CaptainObvious No as there are more than 4 consonants – Ian Miller Dec 7 '15 at 5:01

Consider the layout of the letters given your constraints. You have 3 vowels and 7 consonants. Using 'a' to represent any vowel and 'X' to represent any consonant we have:

aXXaXXaXXX
XaXXaXXaXX
XXaXXaXXaX
XXXaXXaXXa


Then think about possible ways to arrange the different vowels and consonants in those layouts. Watch out for the double T and double R.

EDIT: a bit more generally to think about how to generate the layouts consider this. There are 3 vowels and 7 consonants. The vowels act like gaps between the consonants so there are 2 places which must have 2 consonants. That leaves 3 consonants to place.

• From what you suggested it appears to me I got caught up in tryimg to find a compact formula instead of considering cases. But even with that, let's take case 1. If we take into consideration the double T and R, there should be: $(5)(3)(4)(3)(2)(2)(1)(1)(1)(1)$ but that doesn't appear right to me becuase I didn't really keep positions fixed. – dc3rd Dec 7 '15 at 5:11
• The four layouts don't need to be considered. In each you must place 3 vowels and 7 consonants. So consider the number of ways to do that then multiple by 4. – Ian Miller Dec 7 '15 at 5:14
• Also looking at the numbers you've put down I don't think you've considered the double T and R correctly. – Ian Miller Dec 7 '15 at 5:15
• Extra hint (if you're stuck): You can arrange 7 letters in 7! ways. However if some are the same letter some of those 7! arrangements will be identical. Think about how some of those 7! arrangements could be identical. – Ian Miller Dec 7 '15 at 5:20

Hint:

Another way to think about it: firstly, two routine computations

You can permute the vowels in $3!$ ways.
You can permute the consonants in $\dfrac{7!}{2!2!}$ ways

Now think how many ways you can place the first vowel

You should now surely be able to navigate your way through...

The position of other vowels automatically gets determined, to yield $3!\cdot\dfrac{7!}{2!2!}\cdot 4$

• Based on what the OP has communicated I don't think you can call those routine computations for him. – Ian Miller Dec 7 '15 at 5:46
• Well, OP basically asked for "a hint towards how to consider the constraint." Such a problem is not expected to be given to someone who hasn't been taught to handle permutations of a word with repetitions of letters, and anyway, that, too, was explained in the comment by @Amir earlier. – true blue anil Dec 7 '15 at 6:05

You must fix the position of the vowels such that the nearest vowel is exactly two letters away. As shown pictorially by the other answerer, this gives four possible (generic) arrangements. Once you have this, it's simply $4*$(possible arrangements of vowels)*(possible arrangements of consonants) which is $4*3!*\frac{7!}{2!*2!} = 30240$ since two of the seven consonants are repeated twice.

• Any reason for the downvote? I would certainly love to know if I made a mistake. – YouKnowNothing Dec 7 '15 at 5:32
• I find the answer right, so correcting for downvote given w/o furnishing requested reason (+1) – true blue anil Dec 7 '15 at 19:48