Consider three identical boxes of volume V. the first two boxes will contain particles of two different species 'N' and 'n'. The first box contains 'N' identical non interacting particles in a volume V. The second box contains 'n' non interacting particles. The third box is the result of mixing of these two boxes. Note that the particles are non interacting and the volumes of all the boxes are identical. The total energy which is the just the kinetic energy is conserved. The change in the entropy for such a process is zero from either statistical or thermodynamical calculations. The SMI of the first box is: If we assume that there are M cells in volume V then the number of configurations are M^N(if the particles are distinguishable) and if the particles are indistinguishable then the no. of configurations are M^N / N! hence, SMI ( 1 ) = log M^N / N! similarly, SMI ( 2 ) = log M^n / n! and , SMI ( 3 ) = log M^( N+n ) (distinguishable particles are present) now Δ SMI =SMI(3) - ( SM1( 1 ) + SMI( 2 ) ) = log M^( N+n ) - log M^( N+n ) / N! n! = log N! n! And therefore ΔS = K ln 2 (ΔSMI) which turns out to be not zero. Why is there a discrepancy?. Is there some conceptual mistake in the SMI calculation. In the book, the author explains that the change in SMI is zero since there is no change in volume or energy and hence the locational SMI change and the momenta SMI change are zero...please help.