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Title: Discrete Math
Description: Are you having issues with your Discrete Math course? Well, here's the first week's test answers.

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Assignment Print View

Score:

1
...
14

%

Award: 2 out of 2
...
How many different types of this shirt are made?

72

If there are n1 ways to do the first task and for each of these ways of doing the first task, there are
n2 ways to do the second task, then there are n1n2 ways to do the procedure
...

References
Section Break

http://ezto
...
com/hm
...


4/8/16, 2:58 AM

Award: 2 out of 2
...

If there are n1 ways to do the first task and for each of these ways of doing the first task, there are
n2 ways to do the second task, then there are n1n2 ways to do the procedure
...


http://ezto
...
com/hm
...


4/8/16, 2:58 AM

Award: 2 out of 2
...

If there are n1 ways to do the first task and for each of these ways of doing the first task, there are
n2 ways to do the second task, then there are n1n2 ways to do the procedure
...


http://ezto
...
com/hm
...

References
Section Break

4
...
00 points

Ch 06 Sec 1 Ex 32 1st
Which rule must be used to find the number of strings of eight uppercase English letters that are
there if the letters can be repeated?
The sum rule
The subtraction rule
The division rule
The product rule

By using the rule the number of strings of eight uppercase English letters that are there if the letters
can be repeated is 268 = 208,827,064,576 strings
...
mheducation
...
tpx?todo=printview

Page 4 of 30

Assignment Print View

5
...
00 points

Ch 06 Sec 1 Ex 32 2nd
How many strings of eight uppercase English letters are there if no letter can be repeated?

62990928000

If there are n1 ways to do the first task and for each of these ways of doing the first task, there are
n2 ways to do the second task, then there are n1n2 ways to do the procedure
...
mheducation
...
tpx?todo=printview

Page 5 of 30

Assignment Print View

6
...
00 points

Ch 06 Sec 1 Ex 32 3rd
Which rule must be used to find the number of strings of eight uppercase English letters that start
with X, if letters can be repeated?
The subtraction rule
The sum rule
The product rule
The division rule

By using the rule, there are 267 = 8,031,810,176 strings
...
mheducation
...
tpx?todo=printview

Page 6 of 30

Assignment Print View

7
...
00 points

Ch 06 Sec 1 Ex 32 4th
How many strings of eight uppercase English letters are there that start with X, if no letter can be
repeated?

2422728000

If there are n1 ways to do the first task and for each of these ways of doing the first task, there are
n2 ways to do the second task, then there are n1n2 ways to do the procedure
...
mheducation
...
tpx?todo=printview

Page 7 of 30

Assignment Print View

8
...
00 points

Ch 06 Sec 1 Ex 32 5th
Which rule must be used to find the number of strings of eight uppercase English letters that start
and end with X, if letters can be repeated?
The division rule
The sum rule
The subtraction rule
The product rule

By using the rule, there are 266 = 308,915,776 number of strings
...
mheducation
...
tpx?todo=printview

Page 8 of 30

Assignment Print View

9
...
00 points

Ch 06 Sec 1 Ex 32 6th
How many strings of eight uppercase English letters are there that start with the letters BO (in that
order), if letters can be repeated?

308915776

If there are n1 ways to do the first task and for each of these ways of doing the first task, there are
n2 ways to do the second task, then there are n1n2 ways to do the procedure
...
mheducation
...
tpx?todo=printview

Page 9 of 30

Assignment Print View

10
...
00 points

Ch 06 Sec 1 Ex 32 7th
Which rule must be used to find the number of strings of eight uppercase English letters that start
and end with the letters BO (in that order), if letters can be repeated?
The product rule
The subtraction rule
The division rule
The sum rule

By using the rule, there are 264 = 456,976 number of strings for this case
...
mheducation
...
tpx?todo=printview

Page 10 of 30

Assignment Print View

11
...
00 points

Ch 06 Sec 1 Ex 32 8th

The _____ principle is used to find the number of strings of eight uppercase English letters that
start or end with the letters BO (in that order), if letters can be repeated
...
Therefore by the principle,
the are 266+266−264 = 617,374,576 strings
...
mheducation
...
tpx?todo=printview

Page 11 of 30

Assignment Print View

12
...
00 points

Ch 06 Sec 3 Ex 08
In how many different orders can five runners finish a race if no ties are allowed?

120

The number of different orders is estimated by calculating the value of P(5, 5)
...
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Ch 06 Sec 3 Ex 20
MAIN

Page 12 of 30

Assignment Print View

13
...
00 points

Ch 06 Sec 3 Ex 20 1st
a) exactly three 0s?

120

There are C(10, 3) ways to choose the positions for the 0’s
...


Ch 06 Sec 3 Ex 20
1st

Award: 2 out of 2
...

References
Numeric
Response

Ch 06 Sec 3 Ex 20
2nd

http://ezto
...
com/hm
...


4/8/16, 2:58 AM

Award: 2 out of 2
...

References
Numeric
Response

16
...
00 points

Ch 06 Sec 3 Ex 20 4th
d) at least three 1s?

968

If a string does not have at least three 1’s, then it has 0, 1, or 2 1’s
...

There are 1024 strings in all
...
mheducation
...
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Page 14 of 30

Assignment Print View

4/8/16, 2:58 AM

Ch 06 Sec 3 Ex 27 MAIN
A club has 25 members
...


Ch 06 Sec 3 Ex 27
MAIN

Award: 2 out of 2
...

References
Numeric
Response

Ch 06 Sec 3 Ex 27
1st

http://ezto
...
com/hm
...


4/8/16, 2:58 AM

Award: 2 out of 2
...

References
Numeric
Response

Ch 06 Sec 3 Ex 27
2nd

Ch 06 Sec 3 Ex 30 MAIN
Seven women and nine men are on the faculty in the mathematics department at a school
...
mheducation
...
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Ch 06 Sec 3 Ex 30
MAIN

Page 16 of 30

Assignment Print View

19
...
00 points

Ch 06 sec 3 Ex 30 1st
a) How many ways are there to select a committee of five members of the department if at least one
woman must be on the committee?

4242

There are C(16, 5) ways to select a committee if there are no restrictions and there are C(9, 5) ways
to select a committee from just the 9 men
...


Ch 06 sec 3 Ex 30
1st

Award: 2 out of 2
...

References
Numeric
Response

Ch 06 Sec 3 Ex 30
2nd

http://ezto
...
com/hm
...


4/8/16, 2:58 AM

Award: 2 out of 2
...


1287

The coefficient of x5y8 is


...


Ch 06 Sec 4 Ex 04

Award: 2 out of 2
...


References
Numeric
Response

Ch 06 Sec 4 Ex 06

http://ezto
...
com/hm
...


4/8/16, 2:58 AM

Award: 2 out of 2
...

References
Numeric
Response

Ch 06 Sec 5 Ex 01

http://ezto
...
com/hm
...


4/8/16, 2:58 AM

Award: 2 out of 2
...

References
Numeric
Response

Chapter: 06
Counting

Ch 06 Sec 5
Ex 05

Section: 06
...
mheducation
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Ch 06 Sec 5 Ex 15
MAIN

Page 20 of 30

Assignment Print View

25
...
00 points

Ch 06 Sec 5 Ex 15 1st
a) x1 ≥ 1?

10626

Let
; thus
is the value that
has in excess of its required 1
...

References
Numeric
Response

26
...
00 points

Ch 06 Sec 5 Ex 15 2nd
b)

for i = 1, 2, 3, 4, 5?

1365

Substitute
into the equation for each i; thus is the value that has in excess of its
required 2
...

References
Numeric
Response

Ch 06 Sec 5 Ex 15
2nd

http://ezto
...
com/hm
...


4/8/16, 2:58 AM

Award: 2 out of 2
...
The
restriction on
will be violated if
11
...

References
Numeric
Response

Ch 06 Sec 5 Ex 15
3rd

http://ezto
...
com/hm
...


4/8/16, 2:58 AM

Award: 2 out of 2
...
Then the problem is equivalent to counting the
, where
, subject to

the constraints that
and

...
Thus if the number of solutions is counted to
, subtract
the number of its solutions in which
, and subtract the numbers of its solutions in which
,to arrive at the answer
...
Applying the first restriction reduces the equation to
, which
has C(5 + 1 − 1, 1) = C(5, 1) = 5 solutions
...

References
Numeric
Response

Ch 06 Sec 5 Ex 15
4th

http://ezto
...
com/hm
...


4/8/16, 2:58 AM

Award: 2 out of 2
...

References
Numeric
Response
Ch 06 Sec 5
Ex 21

30
...
05
Generalized
Permutations and
Combinations

Award: 2 out of 2
...


Since MISSISSIPPI has 4 indistinguishable objects of type S, 4 indistinguishable objects of
type I, and 2 indistinguishable objects of type M, it has

= 831,600strings
...
mheducation
...
tpx?todo=printview

= 34,650 strings
...


Preview
Design
Code
References
Multiple
Choice

Chapter: 06
Counting

Ch 06 Sec 5
Ex 30

Section: 06
...
mheducation
...
tpx?todo=printview

Page 25 of 30

Assignment Print View

31
...
00 points

Ch 07 Sec 1 Ex 10
What is the probability that a five-card poker hand contains the two of diamonds and the three of
spades?
(Note: Enter the value in decimal format and report it to four decimal places
...
0075

Compute the number of poker hands that contain the two of diamonds and the three of spades
...
To form the rest of the hand, choose 3 cards
from the 50 remaining cards, so there are C(50, 3) hands containing these two specific cards
...
mheducation
...
tpx?todo=printview

Page 26 of 30

Assignment Print View

32
...
00 points

Ch 07 Sec 1 Ex 21
What is the probability that a fair die never comes up an even number when it is rolled six times?
(Note: Enter the value of probability in decimal format and round it to three decimal places
...
016

There are 2 equally likely outcomes for the parity on the roll of a die—even and odd
...

References
Numeric
Response

Ch 07 Sec 1 Ex 21

http://ezto
...
com/hm
...


4/8/16, 2:58 AM

Award: 2 out of 2
...

(Note: Enter the value in decimal format and round it to two decimal places
...
35

If the numbers are chosen from the integers from 1 to n, then there are C(n, 6) possible entries
...
Therefore, since the winning numbers are picked at random, the


...
mheducation
...
tpx?todo=printview

Ch 07 Sec 1 Ex 34
MAIN

Page 28 of 30

Assignment Print View

34
...
00 points

Ch 07 Sec 1 Ex 34 1st
a) no one can win more than one prize
...

References
Multiple
Choice

Chapter: 07 Discrete
Probability

Ch 07 Sec 1
Ex 34 1st

Section: 07
...
mheducation
...
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Page 29 of 30

Assignment Print View

35
...
00 points

Ch 07 Sec 1 Ex 34 2nd
b) winning more than one prize is allowed
...

References
Multiple
Choice

Chapter: 07 Discrete
Probability

Ch 07 Sec 1
Ex 34 2nd

Section: 07
...
mheducation
...
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Page 30 of 30


Title: Discrete Math
Description: Are you having issues with your Discrete Math course? Well, here's the first week's test answers.