Question

Seven persons sit in a row at random. The probability that three persons A, B, C sit together in a
particular order is

a) 3!/7!
b) 3! 5!/7!
c)4!/7!
d)5!/7!​

Answer 1

Answer:

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Answer 2

The required probability is b) $\dfrac{3! 5!}{7!}$.

Explanation:

We have,

The three particular should always be sit together.

∴ The total number of persons = 5

∴ The five persons can be arranged in = 5 !  ways.

The three persons can be arranged = 3 !  ways.

∴ The number of ways  =  5 !  ×  3 !

∴ P( getting a three persons A, B and C sit together in a  particular order) = $\dfrac{3! 5!}{7!}$

Hence, the required probability is b) $\dfrac{3! 5!}{7!}$.