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A store has an introductory sale on 12 types of candy bars. A customer may choose one bar of any five different types and will be charged no more than $1.75. Show that although different choices may cost different amounts, there must be at least two different ways to choose so that the cost will be the same for both choices.

eminem92
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You find the number of possible choices (the number of "pigeons") and the number of possible prices (the number of "pigeonholes"), and see which one is greater.

There are $_{12}C_5 = 792$ choices (choose 5 from a set of 12 with order not mattering). The number of possible total prices is at most 176, in one-cent increments from \$0.00 up to \$1.75. So no assignment of prices to the items can yield unique total prices for every selection of choices, by the Pigeonhole Principle.

The_Sympathizer
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  • Thank you for your response . Could you suggest on some sites or book for learning discrete mathematics. – eminem92 Jul 10 '13 at 05:00
  • hi , i have tried the same with another method. Could anyone pls tell me if it is correct. – eminem92 Jul 11 '13 at 03:33
  • Question : A store has an introductory sale on 12 types of candy bars. A customer may choose one bar of any five different types and will be charged no more than $1.75. Show that although different choices may cost different amounts, there must be at least two different ways to choose so that the cost will be the same for both choices. – eminem92 Jul 11 '13 at 03:34
  • Solution : Since there are 12 types, then n=12, no. of choices will be 12C5=792 choices. total no of holes - 175 , then by extension method, [(n-1)/m] + 1 ==> (792-1)/175 + 1 =2. so is it possible. – eminem92 Jul 11 '13 at 05:50
  • here 792-1=791 which when divided by 175 give 0.06 which is approximately 1 and plus 1 gives 2. – eminem92 Jul 11 '13 at 05:51
  • Is it correct??? Please let me know as soon as possible .... – eminem92 Jul 11 '13 at 05:51