First learn that the letters that are not vowels are called "consonants",
not "constants".
"ignominious" has 6 vowels and 5 consonants. There are
3 indistinguishable i's
2 indistinguishable o's
2 indistinguishable n's
In class we considered the rearrangement of the word 'ignominious'.
a) How many rearrangements are possible if the first letter must be an 'i' and the last letter must be an 's'?
i _ _ _ _ _ _ _ _ _ s
That's the number of distinguishable arrangements of the 9 other
letters "gnominiou", which has 2 indistinguishable n's, 2 indistinguishable
o's and 2 indistinguishable i's.
Answer: = 45360
b) How many rearrangements are possible if the first letter must be a consonant? (Be careful - the calculations are not the same for all consonants)
Case 1: g comes first. g _ _ _ _ _ _ _ _ _ _ _
That's the number of distinguishable arrangements of the 10 other
letters "inominious", which has 3 indistinguishable i's, 2 indistinguishable
n's and 2 indistinguishable o's.
= 151200 ways
Case 2: n comes first. n _ _ _ _ _ _ _ _ _ _ _
That's the number of distinguishable arrangements of the 10 other
letters "igominious", which has 3 indistinguishable i's, and 2 indistinguishable o's
= 302400 ways
Case 3. m comes first. Same as case 1, which was 151200
Case 4. s comes first. Same as case 1, which was 151200
3×151200 + 302400 = 756000
c) How many rearrangements are possible if the vowels must always be separated by atleast one consonant?
Since we have 6 vowels and only 5 consonants, the only
configuration possible is to alternate them like this:
VCVCVCVCVCV
We can place the 6 vowels distinguishably ways,
since the 3 i's are indistinguishable and the 2 o's are
indistinguishable.
For each way we can place the 6 vowels, we can place the 5
consonants distinguishably in ways since the 2 n's
are indistinguishable.
So that's = 60×60 = 3600
d) How many rearrangements are possible if the consonants must be kept together?
There are 7 ways the 5 consonants can come together
1. CCCCCVVVVVV
2. VCCCCCVVVVV
3. VVCCCCCVVVV
4. VVVCCCCCVVV
5. VVVVCCCCCVV
6. VVVVVCCCCCV
7. VVVVVVCCCCC
Since the number of ways to place the 11 letters distinguishably
in any given configuration of constants and vowels is the same as in
any other given configuration, we can just use the 3600 ways from part
(c) and multiply it by 7.
Answer 3600×7 = 25200
Edwin