Question 1118511
There are 6 ways to choose the first digit, 5 the second and 4 the third. There are 120 ways.

Half or 60 will be odd.

divisible by 5, must end in 5, so 6 ways to choose first and 5 ways to choose the second or 30 ways.

The third term will be the cube of x^3 and will yield x^9.  The third term has coefficient of 15C3=15*14*13*12!/12!*3*2*1=455
this is 455[(x^3)^3-(2/x)^12]=455 x^9-(455*4096)/x^12