9. Suppose 100 identical tickets for rides are distributed among 40 children at
ID: 3028251 • Letter: 9
Question
9. Suppose 100 identical tickets for rides are distributed among 40 children at a carnival (a) How many ways can the tickets be distributed, if each child receives at least two tickets, and all the tickets are distributed (b) How many ways can the tickets be distributed, if each child receives at least one ticket, and some tickets may be left over? (c) suppose one tickets, and each ticket may be used on child has twelve spend her any of six different rides. How many ways can the child tickets, if she can choose any ride any number of times, and the order of choice is unimportant?Explanation / Answer
a] : Once each child has received their two tickets, there are 20 identical tickets left to be given out to the 40 children, with no restrictions. So that there are 40+20-1C20 = 59C20 ways to do this (it’s the number of solutions to a1 + . . . + a40 = 20 with all ai’s non-negative integers, ai representing the number of tickets in excess of 2 that child i gets); so this is the number of ways of doing the distribution .
b]
Once each child has received their one tickets, there are 60 identical tickets left to be given out to the 40 children, with perhaps not all of them given out, and with no restrictions. Imagine a phantom 41st child; if she represents the tickets that are not being distributed, then the question becomes: in how many ways can 60 indistinguishable objects be distributed among 41 distinguishable people.
That is there are 41+601C 60 = 100 C60 such ways.
c]
This is the same as the number of ways of solving a1 + a2 + a3 + a4 + a5 + a6 = 12 with all ai’s non-negative integers (ai representing the number of times ride i is taken). The number of ways is 6+121 C12 = 17C12
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.