Total number of elements = 6
1) In the given table, 23 should be stored at 23mod10 which is equal to 3.
But because of 3 being occupied, it goes to the next value which is 4. Again because of being occupied it is stored at 5.
Therefore, 23 must be in the sequence after 33 and 44.
2) Similarly considering 64, it should be in the sequence after 44 and 23.
3) 91 and 77 can be anywhere in the sequence. Therefore, the total possible positions of 91 and 77 are (6C2)*2! = 30
4) The remaining 4 positions should be in the following order:
33 44 23 64 or
44 33 23 64
2 combinations are possible.
5) Therefore the number of different sequences that would give the resulting table = 30*2 = 60