As the condition says, the store has two digits { 3 ";  one digit  " 5 ", and 3 digits  " 6 ".


So, in all there are 6 digits at the store;

of them, digit  3  has the multiplicity of 2;

         digit  5  has the multiplicity of 1;

    and  digit  6  has the multiplicity of 3.


The number of all possible different six-digit arrangements of these digits is


     =  = 120.


In the formula, 6! is the number of all possible permutations of 6 different digits;

the factors in the denominator account for repeating arrangements, that occur due to presence of indistinguishable digits.