The number 1, 2, 3, ......., n are arranged in a random order. The probabiltiy that the digits 1, 2, 3, .....k (k

Question

The number 1, 2, 3, ......., n are arranged in a random order. The probabiltiy that the digits 1, 2, 3, .....k (k < n) appears as neighbours in that order is (A) (1/n!) (B) (k!/n!) (C) ((n - k)!/n!) (D) ((n - k + 1 )!/n!)

Tardigrade Admin · Accepted Answer

Exhaustive number of cases = n! Assuming the set of numbers 1, 2, 3, ..., k as one the favourable cases = (n - k + 1)! ∴ Probability = ((n - k + 1)!/n!)

Q. The number 1, 2, 3, ......., n are arranged in a random order. The probabiltiy that the digits 1, 2, 3, .....k (k < n) appears as neighbours in that order is

$\frac{1}{n !}$

$\frac{k !}{n !}$

$\frac{( n - k )!}{n !}$

$\frac{( n - k + 1 )!}{n !}$

Exhaustive number of cases = $n!$
Assuming the set of numbers {1, 2, 3, ..., k}
as one the favourable cases = (n - k + 1)!
$∴$ Probability $= \frac{( n - k + 1 )!}{n !}$

Q. The number 1, 2, 3, ......., n are arranged in a random order. The probabiltiy that the digits 1, 2, 3, .....k (k < n) appears as neighbours in that order is

n!1​

n!k!​

n!(n−k)!​

n!(n−k+1)!​

Exhaustive number of cases = n! Assuming the set of numbers {1, 2, 3, ..., k} as one the favourable cases = (n - k + 1)! ∴ Probability =n!(n−k+1)!​

$\frac{1}{n !}$

$\frac{k !}{n !}$

$\frac{( n - k )!}{n !}$

$\frac{( n - k + 1 )!}{n !}$

Exhaustive number of cases = $n!$
Assuming the set of numbers {1, 2, 3, ..., k}
as one the favourable cases = (n - k + 1)!
$∴$ Probability $= \frac{( n - k + 1 )!}{n !}$