Show that for any r, n > 1, n! Σ =r". ni+n2+...+nr=n, ni 20 ni!n2! ...Nr! The summation above runs over all (n1, n2 ...,
Posted: Sun Oct 03, 2021 12:55 pm
Show that for any r, n > 1, n! Σ =r". ni+n2+...+nr=n, ni 20 ni!n2! ...Nr! The summation above runs over all (n1, n2 ..., Nr) satisfying n1 + N2 + ... + ny = n, ni > 0.