" 7.lenm "391682" creg "c_(p)=c_(y)=R

Prove C_(p) - C_(v) = R

Assertion (A) : ""_(n)P_(r) gt ""_(n)C_(r) Reason ""_(n)P_(r) = ""_(n)C_(r) xx r!

In Mayer's relation: C_(P)-C_(V)=R 'R' stands for:

In Mayer's relation: C_(P)-C_(V)=R 'R' stands for:

If x, y and r are positive integers, then ""^(x)C_(r)+""^(x)C_(r-1)+""^(y)C_(1)+""^(x)C_(r-2)""^(y)C_(2)+......+""^(y)C_(r)=

If ""^(n)P_(r)=""^(n)P_((r+1)) and ""^(n)C_(r) = ""^(n)C_(r-1), then (n,r) =

The value of sum_(r=0)^(n) sum_(p=0)^(r) ""^(n)C_(r) . ""^(r)C_(p) is equal to

If x = a - (1)/(a), y = b - (1)/(b) and z = c - (1)/(c ) and a, b, c are in G.P., then show that (x+z)/(y) = r + (1)/(r ) , where r is the common ratio of the G.P.