Question

1. Multiplication theorem There are 20 toys and 12 children. We want to distribute the toys to the children. In each of the cases below, count how many ways can we distribute the toys. (Hint: think, what are my boxes and how can I fill them?). Please show your work. You can leave the expressions as is, so no nieed to do the calculations (a) If cach child gets cxactly one toy. b) Every toy is given out and it is not required that every child gets a toy. It could happen that we give all toys to the same child. (c) We now have a large supply of each of the 20 toys. Each child can get at most one of each toy.

Answer 1

In first part if the toys are indistinguishable then there won't be multiplication with 12! otherwise multiplication will take place.

a)������ Choosing 12 boys (for 12 children) from 20 boys = 20_c_12 They can arrange between themselves is 12! Ways so total no. of ways will be = 20_c_12! b) suppose a_i = i^th student will get a_i number of boys so, a_1 + a_2 + ---- + a_12 = 20 we need to find non-negative integer solution of the equation so the answer is = 20_c_20 + 12 - 1 = 31_c_20 e) Each child gets at most one of each by so each child has two choices 0 and 1 so for each child from 20 boys are have = 2 times 2 times ---- 20 times = 2^20 choices Therefore for 12 children applying multiplication the Orem we have = 2^20 times 2^20 times --- 12 times = (2^20)^12 = 2^240 ways.