Question

The digits 1, 2, 3.... 9 are arranged in a random order, find the probability that 1, 2, 3 will appear as neighbors in the order mentioned.

• A. $\frac{1}{36}$
• B. $\frac{1}{24}$
• C. $\frac{1}{72}$
• D. $\frac{1}{108}$

Answer 1

The correct option is C

$\frac{1}{72}$

The total number of numbers that can be fromed are 9!; because it is the permutation of 9

numbers.

We want 1,2,3 as neighbors in the same order. It means (1,2,3) can be treated as a single

unit. So, there will be a total of 7 digits (9-3+1).

We can have 7! arrangements of such numbers.

$\Rightarrow$ Probability =$\frac{7!}{9!}$

=$\frac{1}{9\times 8}$

=$\frac{1}{72}$