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 :
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
with
no comment