**Concept: The question tests on the basics of Counting**

**Solution:**

Choose the number that has to come twice. This can be done in nC1 = n ways.

Next choose the number that has to come once. This is done in (n-1) C1 = n - 1 ways

These 2 numbers can be arranged in 3! /2! = 3 ways

Total number of ways = n * (n - 1) * 3 = 3n (n - 1) ways

This value has to be between 120 and 130, let us say m.

Then n(n−1)=m/3

From this we understand that m has to be divisible by 3 and m/3 has to be even as n(n - 1) is the product of 2 consecutive numbers which will always be even.

Therefore m = 126 and m/3 = 42

n(n - 1) = 7 * 6. Therefore n = 7

**Option B**