Evaluate `|{:(""^(m)C_(1),""^(m)C_(2),""^(m)C_(3)),(""^(n)C_(1),""^(n)C_(2),""^(n)C_(3)),(""^(p)C_(1),""^(p)C_(2), ""^(p)C_(3)):}|`

Let m, in N and C_(r) = ""^(n)C_(r) , for 0 le r len Statement-1: (1)/(m!)C_(0) + (n)/((m +1)!) C_(1) + (n(n-1))/((m +2)!) C_(2) +… + (n(n-1)(n-2)….2.1)/((m+n)!) C_(n) = ((m + n + 1 )(m+n +2)…(m +2n))/((m +n)!) Statement-2: For r le 0 ""^(m)C_(r)""^(n)C_(0)+""^(m)C_(r-1)""^(n)C_(1) + ""^(m)C_(r-2) ""^(n)C_(2) +...+ ""^(m)C_(0)""^(n)C_(r) = ""^(m+n)C_(r) .

If m,n,r are positive integers such that r lt m,n, then ""^(m)C_(r)+""^(m)C_(r-1)""^(n)C_(1)+""^(m)C_(r-2)""^(n)C_(2)+...+ ""^(m)C_(1)""^(n)C_(r-1)+""^(n)C_(r) equals

Using binomial theorem (without using the formula for sim nC_(r)), prove that ^nC_(4)+^(m)C_(2)-^(m)C_(1)^(n)C_(2)=^(m)C_(4)-^(m+n)C_(1)^(m)C_(3)+^(m+n)C_(2)^(m)C_(2)-^(m+n)C_(3)^(m)C_(1)+^(m+n)C_(4)

Prove that mC_(1)^(n)C_(m)-^(m)C_(2)^(2n)C_(m)+^(m)C_(3)^(3n)C_(m)-...=(-1)^(m-1)n^(m)

(C_(0))^(2)+2(C_(1))^(2)+3(C_(2))^(2)+4(C_(3))^(2)...+(n+1)(c_(n))^(2)