Question

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6. Counting Distributions. (a). How many solutions are there to the equation ni + 12+13 + 14 = 20 if ni, 12, 13, 14 are nonnegative integers and 14 # 4? (b). How many ways are there to give 13 (identical) red pencils and 15 (identical) blue pencils to 11 (different) students? (c). What if each student needs at least 1 pencil of each color?

Answer 1

6. Counting Distributions. (a). How many solutions are there to the equation ni + 12+13 + 14 = 20 if ni, 12, 13, 14 are nonne

solution Given that nithathy tha=20 _o Note: Number of you-negative integer solution of sit xa txat +xm is and is (nem-1 Here basically we have to arrange malt's, and every such arrangements (or) Permutations of those his and mait's Corresponds to Particular solution of a vice versa. so, number of solution of ② is the number of permutations of uits and malts, (nama)! (Aman) nimal) shus unumber of nou negative integer solution liens is 20th 20

is w.if uyah, then with a th3 = 16 so, number solution of 0 when 4454 the number of solution of .. (**a -1) = (13) thus required number of solution is ie, 23 20 16 = 1618 het and be the number of red and blue Pencils given to ith students, for it 1,2,3,..., Xitat. txi so xp, gq are nou-negative integers, and 13 -o [total Red Pencils =13] yityet. +4,1= 15-@[rodal Blue Pencils = 15 Thus, required number of distribute is total number of solution of Ox number of solution of 13+11) * (1516-1) ways to ① x total 13

= (23) -(5) o wote! Number Positive integer solution of xitxa txt... taman is (wmd 621-1) +36-1) +...+(Bem-1) = ham Yityqt...t Ym= nom where y=xq-1 for i=1,21 so yi are nou-negative integer Thus, number of Positive integer solution of eso is the number of nou negative solution of a ter we want set) - ) - () Here we have + xit xa tent x12 13 yit Ya tant Yil = 15 as decribed in ①

distribution Thus required is total number of total number of Postine solution of ② x total number of Positive solution of () (10) 15 X - 11-1 - 14