
#1
Feb1611, 11:19 AM

P: 68

1. The problem statement, all variables and given/known data
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? 2. Relevant equations C(n,k) P(n,k) 3. 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. 



#2
Feb1711, 01:25 AM

HW Helper
P: 3,309

you may wish to consider how to treat repeated letters




#3
Feb1711, 04:33 AM

P: 460

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




#4
Feb1711, 04:41 AM

P: 460

Ways to Arrange the Letters In COMBINATORICS
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 



#5
Feb1711, 05:02 AM

HW Helper
P: 3,309





#6
Feb1711, 05:02 AM

P: 256

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 



#7
Feb1711, 05:03 AM

HW Helper
P: 3,309





#8
Feb1711, 05:40 AM

P: 828





#9
Feb1711, 05:56 AM

P: 828

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 multinomial coefficient. (I prefer binomial in this case.) 


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