Supplementary

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Supplementary Exercisesproblem 1

In the manufacture of a certain type of automobile, fourkinds of major defects and seven kinds of minor defects canoccur. For those situations in which defects do occur, in howmany ways can there be twice as many minor defects as thereare major ones?
Possible ways there can occur twice as many minor defects:
           2 minors and 1 major         4 minors and 2 majors     6 minors and 3 majors
For 2 and 1, 2 of the minors of 7 and 1 of the majors from 4:  C(7, 2) = 21 and C(4,1) =  4.
21*4 = 84 ways to receive 2 minor defects and 1 major.
For 4 minor and 2 major defects: C(7, 4)*C(4, 2) = 35*6 = 210 ways for the 4 minor and 2 major defects.
For 6 minor and 3 major defects: C(7, 6)*C(4, 3) = 7*4 = 28 ways.
The total number of ways to have twice as many minors as majors is by 28 + 210 + 84 = 322

Supplementary Exercisesproblem 2

A machine has nine different dials, each with five settingslabeled 0, 1, 2, 3, and 4.
a)      In how many ways can all the dials on the machine be set?
The machine has 9 dials with 5 settings that are labeled 0, 1, 2, 3, and 4. The process for acquiring the number of different ways of setting in the machine is established by the using n^r for the amount. The n = 5, which is the number of settings and r = 9, which is the number for dials. The formula would be 5^9 = 1953125. The 1953125 is the equal amount of different ways for the use of the 5 sets for the 9 dial numbers in the machine.
b) If the nine dials are arranged in a line at the top of the machine, how many of the         
machine settings have no two adjacent dials with the same setting?
If nine dials are arranged from the top, the formula for the number of ways of arranging the dials without a replication would be by the first dial, which would be by 5^1. This would involve the rest of the dials in line to be represented by 4^8. The form for the total would be (5^1)(4^8) = 327680 ways the nine dials will be arranged without two adjacent dials. The other way to express the math would be by (5^1) = 5 and (4^8) = 4*4*4*4*4*4*4*4. This would then yield 5*4*4*4*4*4*4*4*4 = 327680, which entails the total ways of arranging the dials and not having two adjacent dials with the same setting in the machine.
In how many ways can the letters in WONDERING be arranged with exactly two consecutive vowels?
There are three vowels, which are 0, E, and I. This leaves 6 lettersfrom W, N, D, R, N, and G, which are not vowels but are 6! Then there are 3*2 = 6 ways for arranging the cluster of two vowels. The calculation of 6 letters with 1 vowel for the gap, which would yield 7 and places the two vowels in a cluster. 2 Vowels in a cluster = 7!*2 = 10080

Supplementary Exercisesproblem 10

Mr. and Mrs. Richardson want to name their new daughter so that her initials (first, middle, and last) will be in alphabetical order with no repeated initial. How many such triples of initials can occur under these circumstances?
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EXERCISES 2.1 problem 2

Identify the primitive statements in Exercise 1.
EXERCISES 2.2 problem 4
For primitive statements p, q, r, and s, simplify the compound statement
[[[(p q) r] [(p q)¬r]] ¬q]→s.
EXERCISES 2.5 problem 1
In Example 2.52 why did we stop at 26 and not at 28?
EXERCISES 2.5 problem 2
In Example 2.52, why didn’t we include the odd integers between 2 and 26?
EXERCISES 3.1 problem 1
Which of the following sets are equal?
a) {1, 2, 3} b) {3, 2, 1, 3}
c) {3, 1, 2, 3} d) {1, 2, 2, 3}
EXERCISES 3.1 problem 2
Let A {1, {1}, {2}}. Which of the following statements are true?
a) 1 A     b) {1} A
c) {1} A   d) {{1}} A
e) {2} A   f) {2} A
g) {{2}} A h) {{2}} A
EXERCISES 3.2 problem 3
a) Determine the sets A, B where A − B = {1, 3, 7, 11},
B − A = {2, 6, 8}, and A ∩ B = {4, 9}.
b) Determine the sets C, D where C − D = {1, 2, 4},
D − C = {7, 8}, and C D = {1, 2, 4, 5, 7, 8, 9}.
EXERCISES 3.3 problem 1
During freshman orientation at a small liberal arts college, two showings of the latest James Bond movie were presented. Among the 600 freshmen, 80 attended the first showing and 125 attended the second showing, while 450 didn’t make it to either showing. How many of the 600 freshmen attended twice?