To find the sum of the coefficients in the expansion of (x + y)^n, we can use the binomial theorem. The binomial theorem states that:

(x + y)^n = C(n, 0)x^n + C(n, 1)x^(n-1)y + C(n, 2)x^(n-2)y^2 + ... + C(n, n-1)xy^(n-1) + C(n, n)y^n

where C(n, k) represents the binomial coefficient, which is given by the formula:

C(n, k) = n! / (k!(n-k)!)

Now, we are given that the sum of the coefficients is 4096. This means that:

C(n, 0) + C(n, 1) + C(n, 2) + ... + C(n, n-1) + C(n, n) = 4096

Let's consider each option and try to find the value of the sum of the coefficients for each case:

a) 11C5: The sum of the coefficients would be C(11, 0) + C(11, 1) + C(11, 2) + C(11, 3) + C(11, 4) + C(11, 5) = 1 + 11 + 55 + 165 + 330 + 462 = 1024, which is not equal to 4096.

b) 12C5: The sum of the coefficients would be C(12, 0) + C(12, 1) + C(12, 2) + C(12, 3) + C(12, 4) + C(12, 5) = 1 + 12 + 66 + 220 + 495 + 792 = 1586, which is not equal to 4096.

c) 12C6: The sum of the coefficients would be C(12, 0) + C(12, 1) + C(12, 2) + C(12, 3) + C(12, 4) + C(12, 5) + C(12, 6) = 1 + 12 + 66 + 220 + 495 + 792 + 924 = 2510, which is not equal to 4096.

d) 14C7: The sum of the coefficients would be C(14, 0) + C(14, 1) + C(14, 2) + C(14, 3) + C(14, 4) + C(14, 5) + C(14, 6) + C(14, 7) = 1 + 14 + 91 + 364 + 1001 + 2002 + 3003 + 3432 = 8108, which is not equal to 4096.

Therefore, none of the options a, b, c, and d satisfy the condition that the sum of the coefficients is 4096. So the correct option is none of the above.