Let,
S=0(C130)2+1(C130)2+2(C230)2+3(C330)2..........30(C3030)2
Now rewriting the above expression we get,
S=30(C3030)2+29(C2930)2+28(C2830)2+27(C2730)2..........+0(C030)2
Now adding both the equation we get,
2S=30[(C130)2+(C130)2+(C230)2+(C330)2..........(C3030)2]
⇒S=15[C302×30]
⇒S=15[C3060]=15×(30!)260!
Now comparing with S=(30!)2α60!,
We get α=15