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?
 A: To determine three such tickets, we can pick two numbers $a<b$ in $\{1,\ldots,29\}$, which is possible in $29\choose 2$ ways; then let the tickets be $a,b-a,30-b$. Then do some inclusion and exclusion: Subtract the $14$ cases where $a=b-a$ ($1\le a\le 14$, $b=2a$); subtract the $14$ cases where $a=30-b$ ($1\le a\le 14$, $b=30-a$); subtract the $14$ cases where $b-a=30-b$ ($16\le b\le 29$, $a=2b-30$). Then add back (twice) the one case where $a=b-a=30-b$ ($a=10$, $b=20$). After this, we have counted each (unordered) tripel $\{x,y,z\}\subset\{1,\ldots,30\}$ of ticktes $3!$ times (namely as $x=a,y=b-a,z=30-b$ and all other permutations). Hence the number we look for is is $$\frac{{29\choose 2}-14-14-14+2}6=61. $$
If the tickets actually range from $0$ to $30$ (not $1$ to $30$) the calculation runs like this: There are ${31\choose 2}+31$ ways to pick two numbers $a\le b$ in $\{0,\ldots,30\}$; again let the tickets be $a,b-a,30-b$. Exclude the $16$ cases with $a=b-a$ ($0\le a\le 15$, $b=2a$), the $16$ cases with $a=30-b$ ($0\le a\le 15$, $b=30-a$), the $16$ cases with $b-a=30-b$ ($15\le b\le 30$, $a=2b-30$), and add back $a=10,b=20$ twice. As above divide by $3!$ to obtain
 $$\frac{{31\choose 2}+31-3\cdot 16+2}6=75. $$
In the first interpretation, none of the answer suggestions is true, under the second, B and C are true.
A: 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) 
