Hey,

        I did a mistake last time. Missed out some possibilities.

This time I think the correct answer is 734400

 

This is how I solved it:-

Let the people having five bucks be a1,a2,a3,a4,a5 and those having ten bucks be b1,b2,b3,b4,b5.

Now, we know that the first person must have five rupees and the last one must have ten rupees.

So, first arrange the fivers. there are 5! ways of doing it.

Then put a tenner at the end. There are 5 ways of doing this.

let one such arrangement be:-

a1   a2   a3   a4   a5    b5.

there are five spaces in between them, where the other four tenners have to be arranged.

this can be done in this way:-

possibilities                                                       total ways

 

one person per space                                          ⁵c₄x4!=120

two together and the other two seperate                ⁴c₁x⁴c₂x4!=576

two together and other two together                      ⁴c₁x²c₁x4!=192

three together                                                      ³c₁x⁴c₁x4!=288

four together                                                          ²c₁x4!    =48

Therefore, total possibile ways are 120+576+192+288+48=1224

So,taking into account all possible arrangements of fivers and the last tenner, the final answer is 120x5x1224=734400