First find the non conflicting keys that directly hash to their hash values ...91, 33, 44 and 77
the remaining two 23 and 64 conflict with 33 and 44 in one of the orders like,
91,77,33,44,23,64(this is one sequence possible)
now fix the keys which must be inserted before (that can,t be rearranged)
_33_44_23_64_
keys in bold must not be interchanged, the rest 2 keys(91 and 77) can appear anywhere.
to solve this let us use the simple formula : total permutation /permutation of invalid combinations
total permutation for 6 keys is = 6!
invalid combination is(that should fixed) = 4! //For 33,44,23,64
=> 6!/(4!)
solving this you get 30.
now 44 and 33 can be rearranged for which multiply the result with 2!.
Hence, total number of insertion sequences possible is 60.
