# How many ways INSTRUCTOR can be arranged such that it has two consecutive vowels

How many ways $INSTRUCTOR$ can be arranged such that it has two consecutive vowels?(Three consecutive vowels are not counted i.e $V1V2V3$ is not counted).

I was trying to solve this problem in a different way. But I got a mismatch in the answer. Following is my solution

INSTRUCTOR has three vowels $I,U$ and $O$.

Lets arrange the three vowels like this

_ V1V2 _ V3 _

The above representation shows V1,V2,V3 as vowels and "_" as consonants.

We have total 7 consonants which can fill all these three holes(refereeing to "_").

There must be at least one consonant between V1V2 and V3, else three vowels will come together.

It can be written like this:

$X_1+X_2+X_3=7$

$X_1\ge0,X_2\ge1,X_3\ge0$

This will give result $C(8,2)=28$

We can arrange these 3 vowels in $3!$ ways. Again we can also interchange the position of $V_1,V_2$ and $V_3$ like this.

_V3_V1V2_

So we have total $(28)(3!)(2)$ ways of arranging three vowels. Now the other 7 consonants can be arranged in $\frac{(7!)}{(2!)(2!)}$ ways.

So answer should be $$\frac{(28)(3!)(2)(7!)}{(2!)(2!)}$$

Given answer $$\frac{(64)(3!)(7!)}{(2!)(2!)}$$

What's wrong here?

Help appreciated :)

• How about $V_1V_2V_3$? Mar 24, 2016 at 13:52
• The difference is $8\cdot3! 7!/(2! 2!)$ which is the number of ways of arranging them with the three vowels adjacent. Mar 24, 2016 at 13:58
• Pl. clarify whether $3$ vowels together are to be considered or not. The wording isn't clear to me. Mar 24, 2016 at 14:03
• @QuangHoang 3 vowels(V1V2V3) not considered. Mar 24, 2016 at 14:06

The only thing wrong is that the given answer includes arrangements with all three vowels consecutive. In other words, either you misread the problem, or else the person who came up with the given answer misread it (or forgot that vowel triplets were to be excluded). Your solution to the problem as you read it is correct.

Just for completeness, here's a somewhat simpler way to arrive at the answer for the problem as you read it: There are $7!\over2!2!$ ways to arrange the $7$ consonants. For each of these, there are $8$ places to put a vowel doublet, leaving $7$ places for the vowel singlet. Finally, there are $3!$ ways to pick a doublet, for a total of

$${7!\over2!2!}\cdot8\cdot7\cdot3!={56\cdot3!\cdot7!\over2!2!}$$

arrangements.

To get the "given" answer, we simply count the total number of ways to arrange the $10$ letters and then subtract the number of ways in which the $3$ vowels are all isolated:

$${10!\over2!2!}-{7!\over2!2!}\cdot8\cdot7\cdot6=(10\cdot9-7\cdot6)\cdot8\cdot{7!\over2!2!}=48\cdot8\cdot{7!\over2!2!}={64\cdot6\cdot7!\over2!2!}$$

Let me do my way.

Revised as per clarification of OP

• Group any two vowels together as a flippable object in $\binom32 \times 2 = 6$ ways

• We now have $8$ letters excluding the third vowel. Permute them in $\dfrac{8!}{2!2!}= 8\cdot\dfrac{7!}{2!2!}$ ways

$\uparrow\bullet\uparrow\bullet\uparrow\bullet\circ\bullet\uparrow\bullet\uparrow\bullet\uparrow\bullet\uparrow$ with the $\circ$ representing the $2$ clumped vowels

• Insert the third vowel in $7$ ways at the uparrows

• Putting everything together, we get $56\cdot3!\cdot\dfrac{7!}{2!2!}\cdot$

• -1 That is far too low. You seem to be assuming that the two vowels all come at the start before all the consonants. Mar 24, 2016 at 14:00
• @almagest: No, there was no such assumption. There was a miscount that has been corrected. Mar 24, 2016 at 17:13