<s>[INST] There are 30 tokens numbers from 0 to 30. Find the number of ways of choosing 3 tickets such that the sum of the numbers on the tokens is 30 Then the number of solutions is divisible by 
(A) 2 (B) 3 (C) 5 (D) 7
One way to solve its by finding the number of unequal integral solutions of $x+y+z=30$. 
The possible cases are $x<y<z, y<x<z,..$ (3! Ways)
If $x<y<z$, let $x=a, y=a+b$ and $z=a+b+c$
Then $3a+2b+c=30$ where $0\le a,b,c\le 30$
The number of solutions is the coefficient of $x^{30}$ in $(1-x^3)^{-1}(1-x^2)^{-1}(1-x)^{-1}$ which is as difficult as counting the numbers manually. 
Total number of ways would be $3!$ times the number obtained from above. Hence divisible by 2 and 3. Is this correct?
 [/INST] start|range|combos of 2 to get 30 (Gauss method)
0 .....1-29 ...$\lfloor\frac{29}{2}\rfloor = 14$
1 .....2-27 ...$\lfloor\frac{26}{2}\rfloor = 13$
2 .....3-25 ..$\lfloor\frac{23}{2}\rfloor = 11$
3 .....4-23 ...$\lfloor\frac{20}{2}\rfloor = 10$
4 .....5-21 ...$\lfloor\frac{17}{2}\rfloor = 8$
5 .....6-19 ...$\lfloor\frac{14}{2}\rfloor = 7$
6 .....7-17 ...$\lfloor\frac{11}{2}\rfloor = 5$
7 .....8-15 ...$\lfloor\frac{8}{2}\rfloor = 4$
8 .....9-13 ...$\lfloor\frac{5}{2}\rfloor = 2$
9 .....10-11 ...$\lfloor\frac{2}{2}\rfloor = 1$
Total 75 
Ans:(B) and (C) 
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