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Question
Solve the following:
In how many ways can the letters of the word ARRANGEMENTS be arranged? Find the probability that the consonants are together
Solution
The word' ARRANGEMENTS' has 12 letters of which A, R, N and E appear twice
∴ the number of possible arrangements = n(S)
= `(12!)/(2!)^4`
= `(12!)/16`
Let D ≡ the event that the consonants are together. There are 8 [R, R, N, G, M, N, T, S] consonants and 4 (A, A, E, E) vowels
The 8 consonants can be arranged in `(8!)/(2!)^2 = (8!)/4` ways since R, N appear twice.
Consider consonants as 1 group, then we have to arrange this group and vowels A, A, E, E.
This can be done in `(5!)/(2!)^2` = 30 ways
since A, E appear twice
∴ n(D) = `30 xx (8!)/4`
∴ P(D) = `("n"("D"))/("n"("S"))`
= `(30 xx (8!)/4)/(((12!)/16))`
= `(16 xx 30 xx 8!)/(4 xx 12 xx 11 xx 10 xx 9 xx 8)`
= `1/99`.
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