Recently ,in educational round 83. In question D , i deduced a formula

for(i=n-1;i<=m;i++) { ans=ans+((nCr(i-1,n-2,mod))%mod; } And after that ans=(ans*pow(2,n-3,mod)*(n-2))%mod;

After looking for correct solution i found correct ans was nCr(m,n-1)*(n-2)*pow(2,n-3).

But later observe that for smaller test case my formula was working ,but giving Tle ,at larger tc.

So it that means expression nCr(m,n-1,mod) is equal to summation of nCr(i-1,n-2,mod) ; i range[n-1,m] .

If it is so ,can anyone help me to prove it..

here is my solution link 72867206