Question 1194428: If the permutation of the word WHITE is selected at random, find the probability that the permutation
a. Begins with a consonant
b. Ends with a vowel
c. Has a consonant and vowels alternating
Answer by math_tutor2020(3816) (Show Source):
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Part A
There are 5 letters in the word WHITE
This means there are 5! = 5*4*3*2*1 = 120 permutations possible.
If we want a certain permutation to start with a consonant, then we have these to pick from {W, H, T}. So we have 3 options here.
If the word starts with W, then we have 4! = 4*3*2*1 = 24 different outcomes. The same goes for if the word started with H or T.
This means there are 3*24 = 72 permutations that have a consonant up front.
Divide the number of ways to get what we want (72) over the number of permutations total.
72/120 = (3*24)/(5*24) = 3/5
Answer: 3/5
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Part B
The vowels here are {I, E}
If the word ends in I, then we have 4! = 24 ways to set up the permutation. The same goes for E as well.
2*24 = 48 permutations where it ends with a vowel.
48/120 = (2*24)/(5*24) = 2/5
Answer: 2/5
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Part C
An example where the consonants and vowels alternate would be the "word" WIHET
Due to the fact we have 1 less vowel compared to consonant, this means the vowels must go between the consonants.
This also means the word starts and ends with a consonant.
Let's consider the ways to arrange the three consonants only
3! = 3*2*1 = 6
Now the two vowels
2! = 2*1 = 2
Therefore, we have 6*2 = 12 different ways to have the consonants and vowels alternate
Here are those 12 permutations- HETIW
- HITEW
- HEWIT
- HIWET
- TEHIW
- TIHEW
- TEWIH
- TIWEH
- WEHIT
- WIHET
- WETIH
- WITEH
The first pair has H,T,W in that order with E and I in between those 3 letters.
The second pair has H,W,T and so on.
Each time we move to another pair, apply some kind of swap to the H,W,T.
There are 12 ways to get what we want out of 120 total
12/120 = 1/10
Answer: 1/10
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