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with the provided example, solve the following   A bakery sells chocolate, cinnamon, and plain doughnuts and at a particular time has 8 chocolate, 6 cinnamon, and 5 plain. If a box contains 12 doughnuts, how many different options are there for a box of doughnuts?

with the provided example, solve the following   A bakery sells chocolate, cinnamon, and plain doughnuts and at a particular time has 8 chocolate, 6 cinnamon, and 5 plain. If a box contains 12 doughnuts, how many different options are there for a box of doughnuts?

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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with the provided example, solve the following 

 A bakery sells chocolate, cinnamon, and plain doughnuts and at a particular time has 8 chocolate, 6 cinnamon, and 5 plain. If a box contains 12 doughnuts, how many different options are there for a box of doughnuts?

6. A bakery sells chocolate, cinnamon, and plain doughnuts and at a particular time has 6 chocolate, 6 cinnamon, and 3 plain. If a box contains 12 doughnuts, how many different boxes of doughnuts are possible? Solution. We have the multiset T = {6-a,6-b, 3-c) here a, b, and c are chocolate, cinnamon, and plain doughnuts respectively. Then the number of different boxes of 12 is exactly the num- ber of 12-combinations of the multiset T. Set T = {oa, b, occ). Let S be the set of all 12-combinations of the T. Then |S| = | 12+3-1 12 =. Set A₁ = the set of all 12-combinations in S with at least 7 a, A₂ = the set of all 12-combinations in S with at least 7 b, As = the set of all 12-combinations in S with at least 4 c. Then Ã₁ nÃ₂ nÃ3 is exactly the set of all 12-combinations of T. Note that 5+3-1 | 42| = ( 5+ 3-1) = 5 5 8+3-1 8 |A₁|= |A₂| = |4₁0A3|= Also one sees that A₁ A₂ A3 consists those boxes with with at least 7 a, 7 b, and 4 c. That is impossible. Not by inclusion exclusion principle, we have |A₂A₂A₂|=91- (21+21+45) + (0+3+3) -0 =10. = 21, 2 = 45. 12-7-4+3-1 2 Thus there 10 different boxes. |A₁n4₂| = 0, -¹)=3, 14₂04₂1 = (¹²- = = 21, (12-7- +3-1)= 12-7-4+3-1 2 = 3.
Transcribed Image Text:6. A bakery sells chocolate, cinnamon, and plain doughnuts and at a particular time has 6 chocolate, 6 cinnamon, and 3 plain. If a box contains 12 doughnuts, how many different boxes of doughnuts are possible? Solution. We have the multiset T = {6-a,6-b, 3-c) here a, b, and c are chocolate, cinnamon, and plain doughnuts respectively. Then the number of different boxes of 12 is exactly the num- ber of 12-combinations of the multiset T. Set T = {oa, b, occ). Let S be the set of all 12-combinations of the T. Then |S| = | 12+3-1 12 =. Set A₁ = the set of all 12-combinations in S with at least 7 a, A₂ = the set of all 12-combinations in S with at least 7 b, As = the set of all 12-combinations in S with at least 4 c. Then Ã₁ nÃ₂ nÃ3 is exactly the set of all 12-combinations of T. Note that 5+3-1 | 42| = ( 5+ 3-1) = 5 5 8+3-1 8 |A₁|= |A₂| = |4₁0A3|= Also one sees that A₁ A₂ A3 consists those boxes with with at least 7 a, 7 b, and 4 c. That is impossible. Not by inclusion exclusion principle, we have |A₂A₂A₂|=91- (21+21+45) + (0+3+3) -0 =10. = 21, 2 = 45. 12-7-4+3-1 2 Thus there 10 different boxes. |A₁n4₂| = 0, -¹)=3, 14₂04₂1 = (¹²- = = 21, (12-7- +3-1)= 12-7-4+3-1 2 = 3.
Expert Solution
Step 1

Given:

There are 8 chocolate, 6 cinnamon, and 5 plain doughnuts in this case.

Let,

The number of chocolate doughnuts bought=aThe number of cinnamon doughnuts bought=bThe number of plain doughnuts bought=c

Therefore, 

we need to find the number of non-negative integer solutions of the equation,

a+b+c=12, a≤8, b≤6, c≤5

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