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?
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter10: Matrices
Section10.EA: Extended Application Contagion
Problem 4EA
Related questions
Question
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?
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,
Therefore,
we need to find the number of non-negative integer solutions of the equation,
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