# Ways to Arrange the Letters In COMBINATORICS

## Homework Statement

a) How many arrangements of the letters in COMBINATORICS have no consecutive vowels?
b) In how many of the arrangements in part (a) do the vowels appear in alphabetical order?

C(n,k) P(n,k)

## The Attempt at a Solution

a) First I divided up the consonants and the vowels. My consonants are 2 C's, M, B, N, T, R, and S. My vowels are 2 O's, 2 I's and one A. Now I find the total number of ways to arrange the consonants = 8! Now I have to arrange my vowels such that there are no consecutive vowels. In the diagram below, the K = consonants, and the v's = vowels.

vKvKvKvKvKvKvKvKv

Since there are nine places to place vowels in order to avoid having consecutive vowels, there are C(9,5).

So the solution I am arriving at = 8! * C(9,5) total combinations.

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lanedance
Homework Helper
you may wish to consider how to treat repeated letters

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hey it is not a question to be posted under calculus and beyond

total number of ways to arrange the consonants = 8!/2 as you have one repeated consonant
now why have you done C(9,5).

According to me you can do it using multinomial method. If you do not know what it is, then do not hesitate to ask

lanedance
Homework Helper
So the solution I am arriving at = 8! * C(9,5) total combinations.
You reasoning to this point sounds correct, now you must account for the repeated letters and vowel ordering

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Delta2
Homework Helper
Gold Member
Because the 2Cs are identical the total number to arrange consonants is not 8! but 8!/2!.

Also something similar for vowels cause u have two identical Is and two Os ,you have to divide with 2!2!=4. And u have to multiply by 5! to get a).

For b) you have only one ordering for vowels, that is AIIOO so b) is without the (5!/(2!2!)) factor

lanedance
Homework Helper
total number of ways to arrange the consonants = 8!/2 as you have one repeated consonant
now why have you done C(9,5).

According to me you can do it using multinomial method. If you do not know what it is, then do not hesitate to ask
Maybe you should read over the Original post

hey it is not a question to be posted under calculus and beyond
Yes it is. Combinatorics is most certainly in the realm of calculus and beyond if only for the fact that several ideas in Combo. use ideas from calculus. But, combinatorics is certainly a higher math.

## Homework Statement

a) How many arrangements of the letters in COMBINATORICS have no consecutive vowels?
b) In how many of the arrangements in part (a) do the vowels appear in alphabetical order?

C(n,k) P(n,k)

## The Attempt at a Solution

a) First I divided up the consonants and the vowels. My consonants are 2 C's, M, B, N, T, R, and S. My vowels are 2 O's, 2 I's and one A. Now I find the total number of ways to arrange the consonants = 8! Now I have to arrange my vowels such that there are no consecutive vowels. In the diagram below, the K = consonants, and the v's = vowels.

vKvKvKvKvKvKvKvKv

Since there are nine places to place vowels in order to avoid having consecutive vowels, there are C(9,5).

So the solution I am arriving at = 8! * C(9,5) total combinations.
l would attack this problem similarly, but do something a little different.

First, I like your Kv diagram, I'll use that here.
Second, see how many repeated consonants you have, in this case, there are 2 C's. Now, see how many ways there are to pick two of the K spots to place your two C's. So how many ways can you pick two K spots to place your 2 C's?. Now, you have 6 more K's left and 6 distinct letters with which to fill them, so there are 6! different ways to do this part. Multiply your answer from the C's above and 6! and you know how many ways there are to arrange the consonants.

Now, let's tackle the vowels. First, we note that there are two identical sets of vowels (the two o's and the two i's). There are C(9,2) ways to pick v's to place, say, the o's. Now you have 7 v's left, how many ways are there to pick v's for the i's? Now you how many v's do you have left? How many ways are there to pick a spot for the remaining a?As someone has pointed out, this last part can be answered with the product of three binomial coefficients, or with one multi-nomial coefficient. (I prefer binomial in this case.)