Calculation:
Consider the expression (ry)(N−rn−y)(Nn).
(ry)(N−rn−y)(Nn)=(r!y!(r−y)!)((N−r)!(n−y)!(N−r−(n−y))!)(N!n!(N−n)!)=(r!y!(r−y)!)((N−r)!(n−y)!(N−r−(n−y))!)(n!(N−n)!N!)=(n!y!(n−y)!)(r!(r−y)!)((N−r)!(N−n)!N!(N−r−(n−y))!)=(ny)(r!(r−y)!)((N−r)!(N−n)!N!(N−r−(n−y))!)((N−y)!(N−y)!)
=(ny)×(r!(r−y)!)(N!(N−y)!)×((N−r)!(N−n)!(N−y)!(N−r−(n−y))!)=(ny)×(r!(r−y)!)(N!(N−y)!)×((N−r)!(N−n)!(N−n+(n−y))!(N−r−(n−y))!)=(ny)×(r!(r−y)!)(N!(N−y)!)×((N−r)!(N−r−(n−y))!)((N−n)!(N−n−(n−y))!)
=[(ny)×(r(r−1)(r−2)...(r−y+1)(r−y)!(r−y)!)(N(N−1)(N−2)...(N−y+1)(N−y)!(N−y)!)×((N−r)(N−r−1)...(N−r−(n−y)+1)(N−r−(n−y))!(N−r−(n−y))!)((N−n)(N−n−1)...(N−n−(n−y)+1)(N−n−(n−y))!(N−n−(n−y))!)]=[(ny)×(r(r−1)(r−2)...(r−y+1)N(N−1)(N−2)...(N−y+1))×((N−r)(N−r−1)...(N−r−(n−y)+1)(N−n)(N−n−1)...(N−n−(n−y)+1))]=(ny)×∏a=1yr−y+aN−y+a×∏b=1n−yN−r−n+y+bN−n+b
For value of y being fixed and a belongs to 1,…,y, take the limit with N→∞ for the expression. Also, p=rN is a constant.
limN→∞r−y+aN−y+a=limN→∞r−y+aN−y+a(NN)=limN→∞(r−y+aN)(N−y+aN)=limN→∞(rN−yN+aN)(NN−yN+aN)=(rN−0+0)(1−0+0) (Since rN is constant)=rN=p
For value of y being fixed and b belongs to 1,...,(n−y), take the limit with N→∞ for the expression. If p=rN is a constant, then q=1−p is also a constant.
limN→∞N−r−n+y+bN−n+b=limN→∞N−r−n+y+bN−n+b(NN)=limN→∞(N−r−n+y+bN)(N−n+bN)=limN→∞(NN−rN−nN+yN+bN)(NN−nN+bN)=(1−rN−0+0+0)(1−0+0) (Since rN is constant)=1−rN=q
The value for limN→∞(ry)(N−rn−y)(Nn) is,
limN→∞(ry)(N−rn−y)(Nn)=limN→∞(ny)×∏a=1yr−y+aN−y+a×∏b=1n−yN−r−n+y+bN−n+b=limN→∞(ny)×limN→∞∏a=1yp×limN→∞∏b=1n−yq=limN→∞(ny)pyqn−y
Hence, it is showed that limN→∞(ry)(N−rn−y)(Nn)=(ny)pyqn−y as N→∞.