Checkout Registers and CustomersDate: 02/27/2001 at 23:17:52 From: Phillip Kirkman Subject: Permutations I have two checkout registers, and twenty customers. What formula will find how many different ways I can arrange them? Order does matter. I've tried two checkouts and three people and have 24 different ways I can arrange them. I also tried the same with four people and got 121 ways to arrange them, and five people and 478 ways to arrange them. I don't know what the formula is... Help! Date: 02/28/2001 at 08:05:01 From: Doctor Anthony Subject: Re: Permutations This type of problem can be modelled in the following way: Ten balls are numbered 1,2, ... ,10. In how many ways can these balls be dropped into five different slots, any number into a slot, and with order being important? We let f(10,5) represent the required number of ways. Suppose that a distribution of 9 of the numbers gives f(9,5) possible distributions, and suppose this gives i(1) numbers in box 1, i(2) numbers in box 2 and so on, so that i(1) + i(2) + i(3) + i(4) + i(5) = 9 Then the 10th object can go into box 1 in [i(1)+1] ways box 2 in [i(2)+1] ways and so on. So that [i(1)+1] + [i(2)+1] + ..... + [i(5)+1] = 9 + 5 = 14 ways Since this number is independent of the particular distribution of the nine numbers, we have the relation f(10,5) = 14.f(9,5) = 14.13.f(8,5) = 14.13.12.f(7,5) = 14.13.12.11.f(6,5) = 14.13.12.11.10.f(5,5) = etc, etc, .... = 14.13.12.11.10.9.8.7.6.5 [f(1,5) = 5] = 3632428800 = P(14,10) If m = the number of boxes and n = the number of numbered balls, then the required number of ways = P(m+n-1,n) This can also be written [m]^n = m(m+1)(m+2).....(m+n-1) In the question of the twenty customers and two checkouts, the number of arrangements is P(2+20-1,20) = P(21,20) = 5.109 x 10^19 We can check your work with 3, 4 and 5 customers. With 3 customers P(2+3-1,3) = P(4,3) = 24 agrees with your answer With 4 customers P(2+4-1,4) = P(5,4) = 120 differs by 1 With 5 customers P(2+5-1,5) = P(6,5) = 720 no agreement - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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