The letters of the word ‘EQUATION’ are arranged in a row. Find the

1.	The letters of the word ‘EQUATION’ are arranged in a row. Find the probability that (i) all the vowels are together (ii) arrangement starts with a vowel and ends with a consonant.
Answer» The letters of the word EQUATION can be arranged in 8! ways. ∴ n(S) = 8! There are 5 vowels and 3 consonants. (i) A: all vowels are together we need to arrange (E, U, A, I, O), Q, T, N Let us consider all vowels as one unit. So, there are 4 units, which can be arranged in 4! ways. Also, 5 vowels can be arranged among themselves in 5! ways. ∴ n(A) = 4! × 5! Required probability = P(A) \(\frac {n(A)}{n(S)} = \frac {4!\times5!} {8!} = \frac {1}{14}\) (ii) B: arrangement start with a vowel and ends with a consonant. First and last places can be filled in 5 and 3 ways respectively. Remaining 6 letters are arranged in 6! Ways. ∴ n(B) = 5 × 3 × 6! Required probability = P(B) \(\frac {n(B)}{n(S)} = \frac {5\times3\times6!} {8!} = \frac {15}{56}\)

The letters of the word ‘EQUATION’ are arranged in a row. Find the probability that (i) all the vowels are together (ii) arrangement starts with a vowel and ends with a consonant.

Answer»

The letters of the word EQUATION can be arranged in 8! ways.

∴ n(S) = 8!

There are 5 vowels and 3 consonants.

(i) A: all vowels are together we need to arrange (E, U, A, I, O), Q, T, N

Let us consider all vowels as one unit.

So, there are 4 units, which can be arranged in 4! ways.

Also, 5 vowels can be arranged among themselves in 5! ways.

∴ n(A) = 4! × 5!

Required probability = P(A)

\(\frac {n(A)}{n(S)} = \frac {4!\times5!} {8!} = \frac {1}{14}\)

(ii) B: arrangement start with a vowel and ends with a consonant.

First and last places can be filled in 5 and 3 ways respectively.

Remaining 6 letters are arranged in 6! Ways.

∴ n(B) = 5 × 3 × 6!

Required probability = P(B)

\(\frac {n(B)}{n(S)} = \frac {5\times3\times6!} {8!} = \frac {15}{56}\)

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