To find the value of 'n' in the expansion of (x y)n, where the coefficients of the 4th and 13th terms are equal, we can use the concept of binomial expansion and combinatorics.

Binomial Expansion:
The binomial expansion of (x + y)n can be given by the formula:

(x + y)n = nC0 * xn * y^0 + nC1 * xn-1 * y^1 + nC2 * xn-2 * y^2 + ... + nCr * xn-r * yr + ... + nCn * x^0 * yn

where nC0, nC1, nC2, ..., nCr, ..., nCn are the binomial coefficients, given by the formula:

nCr = n! / (r! * (n-r)!)

where n! denotes the factorial of a number 'n' (n! = n * (n-1) * (n-2) * ... * 2 * 1).

Finding the Coefficients:
To find the coefficients of the 4th and 13th terms, we need to calculate the binomial coefficients for the respective terms.

For the 4th term, we have r = 3:
nC3 = n! / (3! * (n-3)!)

For the 13th term, we have r = 12:
nC12 = n! / (12! * (n-12)!)

Since the coefficients of the 4th and 13th terms are equal, we can set up an equation:

nC3 = nC12

Simplifying the equation:
n! / (3! * (n-3)!) = n! / (12! * (n-12)!)

Cancelling out the common terms:
1 / (3! * (n-3)!) = 1 / (12! * (n-12)!)

Equating the numerators:
1 = (3! * (n-3)!) / (12! * (n-12)!)

Cancelling out the factorials:
1 = (n-3)! / (12! * (n-12)!)

Since the factorials on both sides cancel out, we are left with:
1 = 1 / (12! * (n-12)!)

Simplifying further:
12! * (n-12)! = 1

Since 12! is a large number, the only way the equation can hold true is if (n-12)! = 1.

The only possible value for (n-12)! to be equal to 1 is if (n-12) = 0.

Therefore, n - 12 = 0
n = 12

Hence, the value of 'n' is 12.