Kari Eifler

S

Kari Eifler

(b) To maximize, we use :Corner P=3x+4y(0,0) 0(0,4) 16(3,4) 25(7,1) 25(6,0) 18So the maximum is obtained on the line segment connecting (3,4) and (7,1).That is, it achieves its maximum on the segment y=-3/4t + 25/4 for 3≤t≤7�

Math 141:512 Practice Exam 2, Page 5 of 18 March 4, 2020

6. I’m selling shoes to Company A and B. I want to sell a similar number of shoes to each company,and so the di↵erence between the number of shoes sold to thw two companies cannot di↵er by morethan 10. I only have a maximum of 50 shoes to sell.

I sell the shoes to each company for a profit of $100. How many shoes should I sell to each companyto maximize profit?

Solution:

Since we already know what the variables x and y are, we do not need to define them. The profit isgiven by P = 100x+ 100y = 100(x+ y), subject to the constraints

x� y 10

y � x 10

y + x 50

x � 0

y � 0

and so we get the following diagram:

(0, 0)

(0, 10)

(20, 30)

(30, 20)

(10, 0)x

y

S

The corner points are (0, 0), (0, 10), (20, 30), (30, 20), (10, 0). Then:

Corner Point P = 100x+ 100y(0, 0) 0(0, 10) 1, 000(20, 30) 5, 000(30, 20) 5, 000(10, 0) 1, 000

So it attains it’s maximum along the line connecting (20, 30) and (30, 20).

That is, the solution is (x, y) = (t,�t+ 50) for 20 t 30.

Kari Eifler

|A⋂B| = 10�

Math 141:512 Practice Exam 2, Page 13 of 18 March 4, 2020

22. What are the meanings for P (n, r) and C(n, r)?

Solution:

Suppose we are given n distinct objects and we want r of them.

P (n, r) = n!(n�r)! gives the number of ways to order r of them.

C(n, r) = n!(n�r)!r! gives the number of ways of choose r objects, without regard for order.

23. If C(n, 8) = C(n, 27), what is the value of n?

Solution:

We have

n!

(n� 8)!8!=

n!

(n� 27)!27!

(n� 27)!27! = (n� 8)!8!

Thus we need n� 27 = 8 and 27 = n� 8 which means n = 35.

24. How many ways can we list our three favorite desserts, in order (1st, 2nd, 3rd), from a menu of 10dessert options?

Solution:

Here, the order we list the desserts matters!

We can solve this by P (10, 3) = 10!(10�3)! = 720.

We could also use the multiplication principle:

101st

92nd

83rd

= 720

25. How many ways can we order three di↵erent desserts from a menu of 10 dessert options?

Solution:

Here, the order doesn’t matter! So our answer is just:

C(10, 3) =10!

(10� 3)!3!= 120

Math 141:512 Practice Exam 2, Page 14 of 18 March 4, 2020

26. I am a coin collector and I have coins from 42 di↵erent countries. I bring all my coins to a coincollection meeting with my five friends Alice (A), Bob (B), Charlie (C), David (D), and Eric (E). Iwant to give each of them one of my coins.

How many di↵erent ways can I do this if:

(a) I have brought plenty of coins from each country, and I don’t mind giving more than one friendcoins from the same country.

Solution:

43A

43B

43C

43D

43E

= 435 = 147008443

(b) I have only brought one coin from each country.

Solution:

43A

42B

41C

40D

39E

= P (43, 5) = 115511760

(c) I still have brought only one coin from each country, but only Alice shows up, so I decide to justgive her five coins.

Solution:

C(43, 5) =43!

(43� 5)!5!= 962598

(d) I have only brought 2 identical Canadian coins and 5 identical Euro coins to give my 5 friends.How many ways are there for me to distribute the coins?

Solution:5!

2!3!= 10

Math 141:512 Practice Exam 2, Page 15 of 18 March 4, 2020

27. Suppose I am hanging decorations from 10 nails in my wall and I have 4 identical photo frames, 5identical pieces of artwork, and 1 clock. How many ways could I arrange these items?

Solution:

10!

4!5!1!=

10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 14 · 3 · 2 · 1 · 5 · 4 · 3 · 2 · 1 · 1 = 1260

28. How many ways can I write 8 letter (possibly nonsense) words from the scrabble tiles AADESSST.Note in this case, each letter corresponds to one scrabble tile and so one option would be DESSATAS.

Solution:

8!

2!1!1!3!1!=

8 · 7 · 6 · 5 · 4 · 3 · 2 · 12 · 1 · 1 · 3 · 2 · 1 · 1 = 3360

Math 141:512 Practice Exam 2, Page 16 of 18 March 4, 2020

29. Out of an arrangement of 6 red roses and 5 yellow roses, I want to select 5 roses to give to my mom.I want to include at least 2 red roses. How many options are there?

Solution:

Case 1: 2 red roses (therefore 3 yellow)

C(6,2)red

C(5,3)yellow

= 150

Case 2: 3 red roses (therefore, 2 yellow)

C(6,3)red

C(5,2)yellow

= 200

Case 3: 4 red roses (therefore, 1 yellow)

C(6,4)red

C(5,1)yellow

= 75

Case 4: 5 red roses (therefore no yellow)

C(6,5)red

C(5,0)yellow

= 6

So in total, there are 150 + 200 + 75 + 6 = 431 options.

30. I am packing for a trip and have 3 pairs of jeans, 2 pairs of shorts, and 1 skirt. I want to pack a totalof 4 options, and have at least 2 pairs of jeans. How many ways can I do this?

Solution:

Case 1: 2 pairs of jeans (therefore 2 other options)

C(3,2)jeans

C(3,2)other

= 3 · 3 = 9

Case 2: 3 pairs of jeans (therefore 1 other option)

C(3,3)jeans

C(3,1)other

= 1 · 3 = 3

Case 3: 4 pairs of jeans (therefore no other options)

Note that this case cannot happen since I do not have 4 pairs of jeans to choose from. So we canignore this case.

The total number of ways to pack is 9 + 3 = 12 ways to pack for our trip.

Math 141:512 Practice Exam 2, Page 17 of 18 March 4, 2020

31. In how many di↵erent ways can the letters of the word ’JUDGE’ be arranged such that the vowelsalways come together?

Solution:

The word ’JUDGE’ has 5 letters. It has 2 vowels (UE) and these 2 vowels should always cometogether. Hence these 2 vowels can be grouped and considered as a single letter. That is, JDG(UE).

Hence we can assume total letters as 4 and all these letters are di↵erent. Number of ways to arrangethese letters is 4! = 4 · 3 · 2 · 1 = 24.

In the 2 vowels (UE), all the vowels are di↵erent. Number of ways to arrange these vowels amongthemselves 2! = 2 · 1 = 2.

Total number of ways is 24⇥ 2 = 48

32. How many 3 digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5and none of the digits is repeated?

Solution:

If we want the number to be divisible by 5, then the last digit must be a 0 or 5. Since we are not giventhe option of 0, the last digit must be a 5. Thus, we are just left to decide which digits (2,3,6,7,9)will be the first two digits. We can use the multiplication principle:

1lastdigit

51stdigit

42nddigit

= 20

33. In how many ways can 10 engineers and 4 doctors be seated at a round table without any restriction?

Solution:

If they were seated in a row the answer would be 14!.

However, since they are seated at a round table, this changes the answer a bit. For example: if seating3 people in a row Alice Bob Charlie would be di↵erent than Bob Charlie Alice, however if that wasthe pattern around a circlular table, then they would be the same (you would just rotate the table by1/3). So with three people, there are 3! = 6 di↵erent ways to seat them in a row (ABC, ACB, BAC,BCA, CAB, CBA) but if seating them around a round table there are only two options: clockwiseABC or ACB.

The way we can think of how to solve this smaller case is: let’s have Alice first choose her seat at thetable. We then have two ways to seat the remaining two people: 2! ways.

So we start by at selecting one person to sit down first, and arrange the other people around thatfixed person. How many ways are there to do this? 13! since there are 13 people left to seat.

This is too di�cult to be on an exam, so don’t worry if you’re a bit confused!

34. In how many ways can 10 engineers and 4 doctors be seated at a round table if all the 4 doctors sittogether?

Solution:

This is even more challenging than the previous question!

Since all the 4 doctors sit together, group them together and consider as a single doctor.

Hence, n = totalnumberofpersons = 10 + 1 = 11

These 11 persons can be seated at a round table in (11?1)! = 10! ways.

However these 4 doctors can be arranged among themselves in 4! ways.

Then by the multiplication principle, the required number of ways is 10! · 4!