Question

using the principle of inclusion-exclusion, find the number of solutions of the equation u1+u2+...+u6 = 15, where ui<6,i=1,..,6.

Answer 1

Solution let U1+42 +Uz + ... +46=15 where di 46 let N = number of nonnegabue solution without any condihion, This is because number of Wuhon is - N= ( 1576-1) equivalent do duhbution of 15 15 J= Il identical objects in 6 boxes let Ai denote the set of integer solutions with Ui>6 NlAi) = 946-1 for i=1,2,3,--6 This is because ith box already ( conduins 6 objects) N (ANG) = (346-4) - (3) dr all it) N (Ainaj nau) = 0 Therefore now number of integer solution with Hic 6 is NCÃ Ar nas n-nÃo) Therhore by using Inclusion Exclusion formula Number of solubons = N Ë NCAI) + EN CAIOA;) - EN (Aingna.) t. + NC AIDA --- As) = (49)-6113+ (9 (3) -0---+0 #j - 15504 - 682002 + 15656 = 15504 - 120127 840 - 14332 Thank you