A total of 2016 tickets, numbered 1, 2, 3, 4, ... , 2015, 2016, are placed in an empty bag.

Alberto removes ticket $x$ from the bag. Beck then removes ticket $y$ from the bag. Finally, Chris removes ticket $z$ from the bag.

They notice that $x<y<z$ and $x+y+z=6000$ . In how many ways could this happen?

The answer is 192.

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x+y+z =6000

X+Y+Z=48 ( taking X as 2016- x same has been done for y and z )

Which gives u 50c2 =1225

Now cases where X and Yare same

2X + Z =48 ( 25 cases where Z is an even no till 48)

( similarly we have 2 more cases when X and Z are equal and when Y and Z are equal)

Also the case where all X,Y,Z are same (16 each, but this case has been counted 3 times so we need to balance hence

1225-25*3 +2 = 1152

In this all 3 are unequal and one sixth of the cases will give u when. X< Y<Z will happen

Hence 1152/6= 192

Hence the solutiom