Solution to Math Problem: Choosing Five Fruit Pieces with One Apple from a Basket, Quizzes of Discrete Mathematics

The solution to a problem about calculating the number of ways to choose five pieces of fruit from a basket containing oranges, apples, pears, and grapefruits, with exactly one apple. The solution involves using the concept of combinations with repetitions.

Typology: Quizzes

2010/2011

Uploaded on 06/27/2011

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Math 213, Section B1 (Solution), Quiz 5; February 15, 2008
1.
A basket contains oranges, apples, pears and grapefruits. How many
ways of choosing five pieces of fruit are there, so that exactly one of
the pieces chosen is an apple?
Provide a detailed explanation of your answer.
Solution.
To choose five pieces of fruit so that exactly one is an apple we have
to pick an apple from a basket and put it aside (there is exactly one
way to do that) and then pick four pieces of fruit out of oranges, pears
and grapefruits.
Therefore the number we are asked to find is equal to the number of
ways to choose r= 4 pieces of fruit out of the basket containing n= 3
types of fruit: oranges, pears and grapefruits. This number is equal
to the number of 4-combinations with repetitions out of a set with 3
elements, that is, it is equal to:
C(r+n1, n 1) = C(4 + 3 1,31) = C(6,2) = 6·5
1·2= 15.
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Math 213, Section B1 (Solution), Quiz 5; February 15, 2008

A basket contains oranges, apples, pears and grapefruits. How many ways of choosing five pieces of fruit are there, so that exactly one of the pieces chosen is an apple? Provide a detailed explanation of your answer.

Solution. To choose five pieces of fruit so that exactly one is an apple we have to pick an apple from a basket and put it aside (there is exactly one way to do that) and then pick four pieces of fruit out of oranges, pears and grapefruits. Therefore the number we are asked to find is equal to the number of ways to choose r = 4 pieces of fruit out of the basket containing n = 3 types of fruit: oranges, pears and grapefruits. This number is equal to the number of 4-combinations with repetitions out of a set with 3 elements, that is, it is equal to:

C(r + n − 1 , n − 1) = C(4 + 3 − 1 , 3 − 1) = C(6, 2) =

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