Question : The numbers 1, 2, 3…, n are arrange in random order. The probability that the digits 1, 2, 3.., k, (k<1) appears as neighbours in that order is

(A) 1/n!

(B) k!/n!

(C) (n-k)!n!

(D) (n-k+1)!n!

Answer : option (D)

Solution :

with
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Question : The numbers 1, 2, 3…, n are arrange in random order. The probability that the digits 1, 2, 3.., k, (k<1) appears as neighbours in that order is

(A) 1/n!

(B) k!/n!

(C) (n-k)!n!

(D) (n-k+1)!n!

Answer : option (D)

Solution :