The answer given by Himanshu Pandey in the book Advanced Problems in Organic Chemistry, Chapter Isomerism, Page 76 is 10. But by calculating using the fundamental principle of counting and by viewing a similar question on this site, I figured the answer to be 16. I don't get how he given 10?
@Ivan Neretin, in his comment soon after your post, essentially solved the problem for you. In a comment I subsequently tried to set you in the right direction. Now that the ten structures have been posted in an answer and you have had time to think about a solution, here is the methodology to get the correct answer if you haven't solved it on your own.
All of the 16 permutations are drawn as shown in the chart below. The 10 isomers are in black and the 6 duplicates are in red. Within each group, a red assignment has an identical black one. As an example, in the second group of four, RRSR = RSRR owing to the symmetry in the compound. That is, 2-3-5-6 = 6-5-3-2. The central group of six has "four meso structures", two of which are duplicates.
Note that the number of possible permutations in each group are the coefficients of each term in the expansion of (R + S)4.
The compound 2,3,5,6-tetrachlorooctane in the link does not have a meso compound. The terminal groups, methyl and ethyl are different and therefore meso form is not possible.
The possible combinations have been given below, which are 16.