1. Subsets, Strings, Equations Handout
NAME__________________________________________________ 1.
A vending machine advertises Reese’s Cups, Hershey Bars, Snicker’s Bars, Milky Way Bars, and Granola Bars, and each choice in the vending machine costs $0.50.
Use this class period to work on #1 on this handout, which must be submitted for homework either at the end of this class or in the class indicated on the course schedule.
As you work on each problem, check to see if your final answer is correct. We will go over in class any problems there are questions about. There will also be some time this period for questions.
Quiz #4 COMING UP!
Be sure to do the review problems for this, quiz posted on the internet. The link can be found in the course schedule.
11/29/2018
MATH 106, Section 10

2. CHECK YOUR ANSWERS: 1. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l)
(m) (n) (o)
(p) (q) (r)
(s) (t) (u)
(v)
126
4845 1
1365 56
969 10
560 15
2380
121
3025
111
3844 15
2050 5
425
1610
480
164
179,180
11/29/2018
MATH 106, Section 10

3. Subsets, Strings, Equations Handout
NAME__________________________________________________ 1.
(a)
A vending machine advertises Reese’s Cups, Hershey Bars, Snicker’s Bars, Milky Way Bars, and Granola Bars, and each choice in the vending machine costs $0.50.
How many different ways can Jane spend $2.50 at the vending machine?
r + h + s + m + g = 5
non-negative integers 9!
——– = 126
4! 5!
11/29/2018
MATH 106, Section 10

4. How many different ways can Sam spend $8.00 at the vending machine?
(b)
How many different ways can Sam spend $8.00 at the vending machine?
r + h + s + m + g = 16
non-negative integers
20!
——– = 4845
4! 16!
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MATH 106, Section 10

5. 1  r & 1  h & 1  s & 1  m & 1  g OR positive integers
1.-continued
(c)
How many different ways can Jane spend $2.50 at the vending machine, if she wants to guarantee to purchase at least one of each choice?
r + h + s + m + g = 5
1  r & 1  h & 1  s & 1  m & 1  g OR positive integers
After giving 1 unit to each variable, we restate the problem:
r + h + s + m + g = 0 .
non-negative integers
There is only 1 (one) possible solution.
11/29/2018
MATH 106, Section 10

6. 1  r & 1  h & 1  s & 1  m & 1  g OR positive integers
(d)
How many different ways can Sam spend $8.00 at the vending machine, if he wants to guarantee to purchase at least one of each choice?
r + h + s + m + g = 16
1  r & 1  h & 1  s & 1  m & 1  g OR positive integers
After giving 1 unit to each variable, we restate the problem:
r + h + s + m + g = 11 .
non-negative integers
15!
——– = 1365
4! 11!
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MATH 106, Section 10

7. non-negative integers 8! ——– = 56 3! 5!
1.-continued
(e)
How many different ways can Jane spend $2.50 at the vending machine, if she does not want to purchase any Granola Bars?
r + h + s + m = 5
non-negative integers 8!
——– = 56
3! 5!
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MATH 106, Section 10

8. non-negative integers 19! ——– = 969 3! 16!
(f)
How many different ways can Sam spend $8.00 at the vending machine, if he does not want to purchase any Granola Bars?
r + h + s + m = 16
non-negative integers
19!
——– = 969
3! 16!
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MATH 106, Section 10

9. After giving 3 units to “g”, we restate the problem:
1.-continued
(g)
How many different ways can Jane spend $2.50 at the vending machine, if she wants to purchase exactly three Granola Bars?
r + h + s + m + g = 5
g = 3
After giving 3 units to “g”, we restate the problem:
r + h + s + m = 2
non-negative integers 5!
——– = 10
3! 2!
11/29/2018
MATH 106, Section 10

10. After giving 3 units to “g”, we restate the problem:
How many different ways can Sam spend $8.00 at the vending machine, if he wants to purchase exactly three Granola Bars?
r + h + s + m + g = 16
g = 3
After giving 3 units to “g”, we restate the problem:
r + h + s + m = 13
non-negative integers
16!
——– = 560
3! 13!
11/29/2018
MATH 106, Section 10

11. After giving 3 units to “g”, we restate the problem:
1.-continued
(i)
How many different ways can Jane spend $2.50 at the vending machine, if she wants to purchase at least three Granola Bars?
r + h + s + m + g = 5
g  3
After giving 3 units to “g”, we restate the problem:
r + h + s + m + g = 2
non-negative integers 6!
——– = 15
4! 2!
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MATH 106, Section 10

12. After giving 3 units to “g”, we restate the problem:
(j)
How many different ways can Sam spend $8.00 at the vending machine, if he wants to purchase at least three Granola Bars?
r + h + s + m + g = 16
g  3
After giving 3 units to “g”, we restate the problem:
r + h + s + m + g = 13
non-negative integers
17!
——– = 2380
4! 13!
11/29/2018
MATH 106, Section 10

13. We shall use the GOOD = ALL – BAD principle.
1.-continued
(k)
How many different ways can Jane spend $2.50 at the vending machine, if she wants to purchase at most three Granola Bars?
r + h + s + m + g = 5
g  3
We shall use the GOOD = ALL – BAD principle.
number of solutions in non-negative integers 9!
——– = 126
4! 5!
number of solutions in non-negative integers with g  4 5!
——– = 5
4! 1!
The desired number of solutions is
126 – 5 = 121
11/29/2018
MATH 106, Section 10

14. We shall use the GOOD = ALL – BAD principle.
How many different ways can Sam spend $8.00 at the vending machine, if he wants to purchase at most three Granola Bars?
r + h + s + m + g = 16
g  3
We shall use the GOOD = ALL – BAD principle.
20!
——– = 4845
4! 16!
number of solutions in non-negative integers
number of solutions in non-negative integers with g  4
16!
——– = 1820
4! 12!
The desired number of solutions is
4845 – 1820 = 3025
11/29/2018
MATH 106, Section 10

15. We shall use the GOOD = ALL – BAD principle.
1.-continued
(m)
How many different ways can Jane spend $2.50 at the vending machine, if there are only two Milky Way Bars left in the vending machine?
r + h + s + m + g = 5
m  2
We shall use the GOOD = ALL – BAD principle.
number of solutions in non-negative integers 9!
——– = 126
4! 5!
number of solutions in non-negative integers with m  3 6!
——– = 15
4! 2!
The desired number of solutions is
126 – 15 = 111
11/29/2018
MATH 106, Section 10

16. We shall use the GOOD = ALL – BAD principle.
How many different ways can Sam spend $8.00 at the vending machine, if there are only five Milky Way Bars left in the vending machine?
r + h + s + m + g = 16
m  5
We shall use the GOOD = ALL – BAD principle.
20!
——– = 4845
4! 16!
number of solutions in non-negative integers
number of solutions in non-negative integers with m  6
14!
——– = 1001
4! 10!
The desired number of solutions is
4845 – 1001 = 3844
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MATH 106, Section 10

17. We shall use the GOOD = ALL – BAD principle.
1.-continued
(o)
How many different ways can Jane spend $2.50 at the vending machine, if she wants to purchase at least three Granola Bars and there are only two Milky Way Bars left in the vending machine?
r + h + s + m + g = 5
g  3 & m  2
We shall use the GOOD = ALL – BAD principle. 6!
——– = 15
4! 2!
number of solutions in non-negative integers with g  3
number of solutions in non-negative integers with g  3 & m  3
The desired number of solutions is
15 – 0 = 15
11/29/2018
MATH 106, Section 10

18. We shall use the GOOD = ALL – BAD principle.
How many different ways can Sam spend $8.00 at the vending machine, if he wants to purchase at least three Granola Bars and there are only five Milky Way Bars left in the vending machine?
r + h + s + m + g = 16
g  3 & m  5
We shall use the GOOD = ALL – BAD principle.
17!
——– = 2380
4! 13!
number of solutions in non-negative integers with g  3
number of solutions in non-negative integers with g  3 & m  6
11!
——– = 330
4! 7!
The desired number of solutions is
2380 – 330 = 2050
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MATH 106, Section 10

19. We shall use the GOOD = ALL – BAD principle.
1.-continued
(q)
How many different ways can Jane spend $2.50 at the vending machine, if she wants to purchase at most three Granola Bars and at least four Reese’s Cups?
r + h + s + m + g = 5
r  4 & g  3
We shall use the GOOD = ALL – BAD principle. 5!
——– = 5
4! 1!
number of solutions in non-negative integers with r  4
number of solutions in non-negative integers with r  4 & g  4
The desired number of solutions is
5 – 0 = 5
11/29/2018
MATH 106, Section 10

20. We shall use the GOOD = ALL – BAD principle.
How many different ways can Sam spend $8.00 at the vending machine, if he wants to purchase at most three Granola Bars and at least eight Reese’s Cups?
r + h + s + m + g = 16
r  8 & g  3
We shall use the GOOD = ALL – BAD principle.
12!
——– = 495
4! 8!
number of solutions in non-negative integers with r  8
number of solutions in non-negative integers with r  8 & g  4 8!
——– = 70
4! 4!
The desired number of solutions is
495 – 70 = 425
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MATH 106, Section 10

21. We shall use the GOOD = ALL – BAD principle.
1.-continued
(s)
How many different ways can Sam spend $8.00 at the vending machine, if he wants to purchase at least four, but no more than nine, Reese’s Cups?
r + h + s + m + g = 16
4  r  9
We shall use the GOOD = ALL – BAD principle.
16!
——– = 1820
4! 12!
number of solutions in non-negative integers with r  4
number of solutions in non-negative integers with r  4 & r  10
10!
——– = 210
4! 6!
The desired number of solutions is
1820 – 210 = 1610
11/29/2018
MATH 106, Section 10

22. We shall use the GOOD = ALL – BAD principle.
How many different ways can Sam spend $8.00 at the vending machine, if he wants to purchase at least one of each choice and at least four, but no more than nine, Reese’s Cups?
r + h + s + m + g = 16
positive integers 4  r  9
We shall use the GOOD = ALL – BAD principle.
12!
——– = 495
4! 8!
number of solutions in positive integers with r  4
number of solutions in positive integers with r  4 & r  10 6!
——– = 15
4! 2!
The desired number of solutions is
495 – 15 = 480
11/29/2018
MATH 106, Section 10

23. We shall use the GOOD = ALL – BAD principle.
1.-continued
(u)
Right next to the vending machine described previously is a second vending machine advertising a bag of potato chips, a bag of pretzels, a bag of cheese doodles, and a bag of cookies, each bag costing $ How many different ways can Jane spend $2.50 at both vending machines, if she wants to purchase at least two, but no more than four, bags of potato chips?
r + h + s + m + g + c + p + d + k = 5
2  c  4
We shall use the GOOD = ALL – BAD principle.
11!
——– = 165
8! 3!
number of solutions in non-negative integers with c  2
number of solutions in non-negative integers with c  2 & c  5 8!
——– = 1
8! 0!
The desired number of solutions is
165 – 1 = 164
11/29/2018
MATH 106, Section 10

24. We shall use the GOOD = ALL – BAD principle.
(v)
Right next to the vending machine described previously is a second vending machine advertising a bag of potato chips, a bag of pretzels, a bag of cheese doodles, and a bag of cookies, each bag costing $ How many different ways can Sam spend $8.00 at both vending machines, if he wants to purchase at least three, but no more than six, bags of potato chips?
r + h + s + m + g + c + p + d + k = 16
3  c  6
We shall use the GOOD = ALL – BAD principle.
21!
——– = 203,490
8! 13!
number of solutions in non-negative integers with c  3
number of solutions in non-negative integers with c  3 & c  7
17!
——– = 24,310
8! 9!
The desired number of solutions is
203,490 – 24,310 = 179,180
11/29/2018
MATH 106, Section 10