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