12/14/2015math 106, section 101 subsets, strings, equations handout 1. a vending machine advertises...
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04/21/23 MATH 106, Section 10 1
Subsets, Strings, Equations Handout
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.
NAME__________________________________________________
Use this class period to work on this handout, which must be submitted for homework at the end of class.
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.
04/21/23 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)
2. (a) (b) (c)
(d) (e) (f)
(g) (h) (i)
(j) (k) (l)
126 4845 1
1365 56 969
10 560 15
2380 121 3025
111 3844 15
2050 5 425
1610 480 164
179,180
84 0 35
4 7 0
262,144 6765 6765
2 255,379 262,142
04/21/23 MATH 106, Section 10 3
Subsets, Strings, Equations Handout
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?
NAME__________________________________________________
r + h + s + m + g = 5
non-negative integers
9!——– = 126 4! 5!
04/21/23 MATH 106, Section 10 4
(b) How many different ways can Sam spend $8.00 at the vending machine?
r + h + s + m + g = 16
non-negative integers
20!——– = 48454! 16!
04/21/23 MATH 106, Section 10 5
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 integersThere is only 1 (one) possible solution.
04/21/23 MATH 106, Section 10 6
(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!——– = 13654! 11!
04/21/23 MATH 106, Section 10 7
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!
04/21/23 MATH 106, Section 10 8
(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!——– = 9693! 16!
04/21/23 MATH 106, Section 10 9
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!
04/21/23 MATH 106, Section 10 10
(h) 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!——– = 5603! 13!
04/21/23 MATH 106, Section 10 11
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!
04/21/23 MATH 106, Section 10 12
(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!——– = 23804! 13!
04/21/23 MATH 106, Section 10 13
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 3We shall use the GOOD = ALL – BAD principle.
number of solutions in non-negative integers
number of solutions in non-negative integers with g 4
9!——– = 126 4! 5!
5!——– = 5 4! 1!
The desired number of solutions is
126 – 5 = 121
04/21/23 MATH 106, Section 10 14
(l) 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 3We shall use the GOOD = ALL – BAD principle.
number of solutions in non-negative integers
number of solutions in non-negative integers with g 4
20!——– = 48454! 16!
16!——– = 18204! 12!
The desired number of solutions is
4845 – 1820 = 3025
04/21/23 MATH 106, Section 10 15
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 2We shall use the GOOD = ALL – BAD principle.
number of solutions in non-negative integers
number of solutions in non-negative integers with m 3
9!——– = 126 4! 5!
6!——– = 15 4! 2!
The desired number of solutions is
126 – 15 = 111
04/21/23 MATH 106, Section 10 16
(n) 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 5We shall use the GOOD = ALL – BAD principle.
number of solutions in non-negative integers
number of solutions in non-negative integers with m 6
20!——– = 48454! 16!
14!——– = 10014! 10!
The desired number of solutions is
4845 – 1001 = 3844
04/21/23 MATH 106, Section 10 17
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.
number of solutions in non-negative integers with g 3 number of solutions in
non-negative integers with g 3 & m 3
6!——– = 15 4! 2!
0
The desired number of solutions is
15 – 0 = 15
04/21/23 MATH 106, Section 10 18
(p) 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.
number of solutions in non-negative integers with g 3 number of solutions in
non-negative integers with g 3 & m 6
17!——– = 23804! 13!
11!——– = 330 4! 7!
The desired number of solutions is
2380 – 330 = 2050
04/21/23 MATH 106, Section 10 19
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.
number of solutions in non-negative integers with r 4 number of solutions in
non-negative integers with r 4 & g 4
5!——– = 5 4! 1!
0
The desired number of solutions is
5 – 0 = 5
04/21/23 MATH 106, Section 10 20
(r) 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.
number of solutions in non-negative integers with r 8 number of solutions in
non-negative integers with r 8 & g 4
12!——– = 495 4! 8!
8!——– = 70 4! 4!
The desired number of solutions is
495 – 70 = 425
04/21/23 MATH 106, Section 10 21
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.
number of solutions in non-negative integers with r 4 number of solutions in
non-negative integers with r 4 & r 10
16!——– = 18204! 12!
10!——– = 210 4! 6!
The desired number of solutions is
1820 – 210 = 1610
04/21/23 MATH 106, Section 10 22
(t) 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.
number of solutions in positive integers with r 4 number of solutions in
positive integers with r 4 & r 10
12!——– = 495 4! 8!
6!——– = 15 4! 2!
The desired number of solutions is
495 – 15 = 480
04/21/23 MATH 106, Section 10 23
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 $0.50. 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.
number of solutions in non-negative integers with c 2 number of solutions in
non-negative integers with c 2 & c 5
11!——– = 165 8! 3!
8!——– = 1 8! 0!
The desired number of solutions is
165 – 1 = 164
04/21/23 MATH 106, Section 10 24
(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 $0.50. 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.
number of solutions in non-negative integers with c 3 number of solutions in
non-negative integers with c 3 & c 7
21!——– = 203,4908! 13!
17!——– = 24,310 8! 9!
The desired number of solutions is
203,490 – 24,310 = 179,180
04/21/23 MATH 106, Section 10 25
2.
(a)
(b)
A child is playing with large box of blocks that are identical, except that some are red and some are blue.
How many different ways can six red blocks and three blue blocks be arranged in a row?
How many different ways can six red blocks and three blue blocks be arranged in a row so that none of the red blocks are adjacent?
9!——– = 84 6! 3!
0
NAME__________________________________________________
04/21/23 MATH 106, Section 10 26
(c)
(d)
How many different ways can six red blocks and three blue blocks be arranged in a row so that none of the blue blocks are adjacent?
How many different ways can six red blocks and three blue blocks be arranged in a row so that all of the red blocks are adjacent?
7!——– = 35 3! 4!
4
04/21/23 MATH 106, Section 10 27
2.-continued (e)
(f)
How many different ways can six red blocks and three blue blocks be arranged in a row so that all of the blue blocks are adjacent?
How many different ways can six red blocks and three blue blocks be arranged in a row so that blocks of the same color are not adjacent?
7
0
04/21/23 MATH 106, Section 10 28
(g)
(h)
How many different ways can 18 blocks, each one either red or blue, be arranged in a row?
How many different ways can 18 blocks, each one either red or blue, be arranged in a row so that none of the blue blocks are adjacent?
218 = 262,144
F(18) = 6765
04/21/23 MATH 106, Section 10 29
2.-continued (i)
(j)
How many different ways can 18 blocks, each one either red or blue, be arranged in a row so that none of the red blocks are adjacent?
How many different ways can 18 blocks, each one either red or blue, be arranged in a row so that blocks of the same color are not adjacent?
F(18) = 6765
2
04/21/23 MATH 106, Section 10 30
(k)
(l)
How many different ways can 18 blocks, each one either red or blue, be arranged in a row so that at least two of the red blocks are adjacent?
How many different ways can 18 blocks, each one either red or blue, be arranged in a row so that at least two blocks of the same color are adjacent?
218 – F(18) = 262,144 – 6765 = 255,379
218 – 2 = 262,144 – 2 = 262,142