Question 1151170: 2. If the permutation of the word WHITE is selected at random, how many of the
permutations
i. Begins with a consonant?
ii. Ends with a vowel?
iii. Has a consonant and vowels alternating?
Answer by jim_thompson5910(35256)   (Show Source):
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i.
Consonants to choose from = {W, H, T}
Vowels to choose from = {I, E}
We have 3 consonants to fill slot 1.
Then we have 4 choices for slot 2
Then 3 choices for slot 3
2 choices for slot 4
1 choice for slot 5
Overall, 3*4*3*2*1 = 3*4! = 3*24 = 72
There are 72 permutations that begin with a consonant.
Answer: 72
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ii.
Slot 5 has 2 choices, either an E or an I.
After slot 5 is filled, we have 4 choices for slot 1
Then 3 choices for slot 2
2 choices for slot 3
1 choice for slot 4
2*4*3*2*1 = 2*4! = 2*24 = 48
There are 48 permutations that have a vowel as the last letter.
Answer: 48
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iii.
define two variable placeholders
c = consonant from the list {W, H, T}
V = vowel from the list {I, E}
We want the consonants c and the vowels V to alternate. If we start with a consonant, then we must have a vowel next, and then another consonant, etc etc.
That pattern would be
cVcVc
We can't start with a vowel because we'd have
VcVcc
we run out of vowels once we reach slot 3.
And of course we can't have something like this
ccVcV
because of the adjacent consonants in slots 1 and 2.
So again we only have something of the form
cVcVc
Consider just the consonants for now. We basically have 3 slots to fill and 3 items to fill them. There are 3! = 3*2*1 = 6 ways to do this. Order matters.
Then we have 2 other slots for the vowels, so 2! = 2*1 = 2 ways to fill the vowel slots.
Overall, there are 6*2 = 12 ways to have a permutation of letters such that the consonants and vowels alternate.
Answer: 12
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