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Chilukuri Sai Kartik,
12 years ago
Grade:12
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1 Answers
abhishek athreya
19 Points
12 years ago
in the word LOGARITHM
the position of vowels is 2,4,6
the position of consonants is 1,3,5,7,8,9
as their relative positions should be maintained
the vowels can occupy only positions 2,4,6
and the consonants can only occupy positions 1,3,5,7,8,9
the permutation among vowels = 3!
the permutation among consonants =6!
thus total permutation = 6!*3!
=4320
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The letters of the word 'LOGARITHM' are arranged at random. The probability that arrangement starts with vowel and end with consonant is ____________.
Options
`1/9`
`[7!]/[9!]`
`18/[9!]`
`1/4`
Solution
The letters of the word 'LOGARITHM' are arranged at random. The probability that arrangement starts with vowel and end with consonant is `underline[1/4]`.
Explanation:
Given, word 'LOGARITHM'
Total number of arrangements = 9!
Here, vowel [O, A, I]
Consonant [L, G, R, T, H, M]
Total number of ways when starts with vowel and end with consonant is `""^3"C"_1 xx ""^6"C"_1 xx 7!`
∴ Required probability = `[""^3"C"_1 xx ""^6"C"_1 xx 7!]/[9!] = 1/4`
Concept: Probability Distribution - Cumulative Probability Distribution of a Discrete Random Variable
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The letters of the word LOGARITHM are arranged at random. Find the probability that start with vowel and end with ends with consonant.
Solution
There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in 9P9 = 9! ways.
∴ n[S] = 9!
Let E be the event that word starts with vowel and ends with consonant.
There are 3 vowels and 6 consonants in
the word LOGARITHM.
∴ The first place can be filled in 3 different ways and the last place can be filled in 6 ways.
Now, remaining 7 letters can be arranged in 7 places in 7P7 = 7! ways
∴ n[E] = 3 × 7! × 6
∴ P[E] = `["n"["E"]]/["n"["S"]`
= `[3 xx 7! xx 6]/[9!]`
Concept: Elementary Properties of Probability
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