How many words can be made from REGULATION where the vowels are in alphabetical order

Hi Gus,

I have the following math problem that I am stuck on:

How many words can be made from REGULATION where all the vowels are in alphabetical order?

I solved it two different ways and got two different answers.

1. I did 10C5 for choosing 5 spots then multiplied by 1 for the one order the vowels can be in. Then I multiplied by 5! for all the rearrangements of the other letters. This is (10 C 5) * 1 * 5! = 30240

2. I recognize the symmetry where half the arrangements, a is before e, half the arrangements e is before i, half the arrangements i is before o, half the arrangements o is before u. This symmetry means that the number of arrangements is 10!/(2^4)

which equals 226800.

Which is correct? Why is the other wrong?

Thanks for the help

Re: How many words can be made from REGULATION with the vowels in alphabetical order?

Quote:

Originally Posted by

**ShadowKnight8702** 2. I recognize the symmetry where half the arrangements, a is before e, half the arrangements e is before i, half the arrangements i is before o, half the arrangements o is before u. This symmetry means that the number of arrangements is 10!/(2^4)

which equals 226800.

No. What about A before I, A before O, A before U, E before O, E before U, I before U? In any case, this is the wrong way of counting the permutations of the vowels. Now, you know that the number of ways the vowels can be permuted is $\displaystyle 5!$. Hence the number of solutions the problem requires is $\displaystyle \frac{10!}{5!}=30240$ as in your first method.