182 Appendix A
(x+y)N?Andthe answeristhe same: zero unlessn+m=N,butifn+m=Nthen
the required number isN!/(n!×m!).
3 .Inhow many ways can theNobjectsbe arrangedif they are now split up intor+ 1
piles withnnjobjects in pile numberj(((j=0, 1, 2,...r)?
The answeris zero unlessN=
∑
nnj.Ifthennjsdo sum toNthen the requirednumber
ofwaysisN!/(n 0 !n 1 !...nr!).
This result follows as a straightforward extension of the proof of problem 2,
equation (A.1)becoming:N!=t×
∏
nnj!.Thisimportant resultisthe one used
extensivelyin Chapter 2.
Again one may note that this problem may be thought of (and solved) as the question
ofthemultinomialtheorem:whatisthe coefficient ofyn 00 yn 11 y 2 n^2 ...inthe expansion
of(y 0 +y 1 +y 2 +···)N? And the answer is identical.