Statistics

Total number of ways to arrange the 6 digits
= 6P6
= 6!
= 720


There are two cases that the first position is occupied by 1,2,3 or 4 and the last position is occupied by 3,4,5 or 6:

Case I : The first position is 1 or 2, the last position is 3, 4, 5 or 6. The rest 4 positions are occupied by the rest 4 digits.
Number of ways to put the 6 digits in case I
= 2P1 x 4P1 x 4P4
= 2 x 4 x 4!
= 192

Case II : The first position is 3 or 4, the last position is 5, 6 or (3 or 4 that does not occupy the first place). The rest 4 positions are occupied by the rest 4 digits.
Number of ways to put the 6 digits in Case II
= 2P1 x 3P1 x 4P4
= 2 x 3 x 4!
= 144

Total number ways that the 6 digits are in required arrangement
= 192 + 144
= 336


Probability that the digits 6 digits are in required arrangement
= 336/720
= 7/15
=