Find the probability of getting R in the first place and M in the last

1.	Find the probability of getting R in the first place and M in the last place when all the letters of the word RANDOM are arranged in all possible ways.
Answer» Here, there are 6 letters R, A, N, D, O, M in the word RANDOM. ∴ The total number of ways of arranging these six letters is, n = ₆P⁶ = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720. A = Event that getting R in the first place and M in the last place. ∴ Favourable outcomes for the event A are obtained as follows: ∴ m = ₁P¹ × ₄P⁴ × ₁P¹ = 1! × 4! × 1! = 1 × 24 ×1 = 24 Hence p(A) = \(\frac{m}{n} = \frac{24}{720} = \frac{1}{30}\)

Find the probability of getting R in the first place and M in the last place when all the letters of the word RANDOM are arranged in all possible ways.

Answer»

Here, there are 6 letters R, A, N, D, O, M in the word RANDOM.

∴ The total number of ways of arranging these six letters is,

n = ₆P⁶ = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.

A = Event that getting R in the first place and M in the last place.

∴ Favourable outcomes for the event A are obtained as follows:

∴ m = ₁P¹ × ₄P⁴ × ₁P¹

= 1! × 4! × 1!

= 1 × 24 ×1 = 24

Hence p(A) = \(\frac{m}{n} = \frac{24}{720} = \frac{1}{30}\)

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Find the probability of getting R in the first place and M in the last place when all the letters of the word RANDOM are arranged in all possible ways.

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