The spin of a particle is a quantum mechanical property that can take on discrete values. In this case, the particles have a spin of 1, which means that their spin can take on three possible values: +1, 0, or -1.

To determine the entropy of the system, we need to calculate the number of microstates, Ω, that correspond to a given macrostate. The macrostate is defined by the total spin of the system, which can range from -N to +N in steps of 2.

The number of microstates for each macrostate can be determined using the formula for the degeneracy of a system with multiple particles and multiple energy levels. In this case, the energy levels correspond to the possible spin values of the particles.

The degeneracy is given by the formula:

Ω = (2J + 1)^(N_up) * (2J + 1)^(N_zero) * (2J + 1)^(N_down)

where J is the spin of the particle (in this case, J = 1), N_up is the number of particles with spin +1, N_zero is the number of particles with spin 0, and N_down is the number of particles with spin -1.

Since the particles are distinguishable, we need to consider the number of ways to arrange the particles among the energy levels. This is given by the multinomial coefficient:

C = N! / (N_up! * N_zero! * N_down!)

The total number of microstates for a given macrostate is then given by Ω * C.

To calculate the entropy, we use Boltzmann's entropy formula:

S = kB * ln(Ω * C)

where kB is Boltzmann's constant.

Substituting the expression for Ω * C, we have:

S = kB * ln[(2J + 1)^(N_up) * (2J + 1)^(N_zero) * (2J + 1)^(N_down) * N! / (N_up! * N_zero! * N_down!)]

Simplifying the expression, we find:

S = N * kB * ln(2J + 1) + kB * ln(N!)

The second term, kB * ln(N!), represents the combinatorial entropy of the system and is typically much smaller than the first term for large N.

Since we are given that the entropy is of the form NkBln x, we can equate the expression above with NkBln x:

N * kB * ln(2J + 1) + kB * ln(N!) = NkBln x

Cancelling out the kB terms and rearranging the equation, we find:

ln(N!) = ln x - ln(2J + 1)

The equation above implies that the value of x should be chosen such that N! is equal to (2J + 1) to the power of N. In this case, J = 1, so (2J + 1) = 3, and x = 3 is the correct answer.