Ma 2201/ cs2022 Test #2 Tuesday, May 2, 2006 No Calculators Allowed. You may leave large calculations unevaluated if the calculation is not on the attached sheet. Show all work!




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MA 2201/ CS2022 Test #2

Tuesday, May 2, 2006
No Calculators Allowed. You may leave large calculations unevaluated if the calculation is not on the attached sheet.

Show all work! (No credit for just numbers or final answers)

#1. a) How many ways can a 5-card poker hand be dealt from 52 cards?
C(52,5) = 52! = 2,598,960

5! * 47!
b) A flush is 5 cards of the same suit. How many 5-card hands contain flushes?
This is the number of suits - 4 times the number of ways to pick 5 cards from a suit – C(13,5)
Number of hands with a flush = 4 * C(13,5) = 4 * 13! = 5148

5! 8!
c) What is the probability of being dealt a flush in a 5-card poker hand?
P(flush) = 5148/2,598,960 = (approx) 0.00198
d) What is the probability of getting a 5-card hand which is not a flush?
P(not a flush) = 1 – 0.00197 = 0.99803
#2. How many ways can four of the letters in the word “beach” be written if the last letter must be “h”?
Since the last letter must be “h”, there are really only three letters to be chosen to precede it (from the remaining four)
P(4,3) = 4!/(4-3)! = 24
#3. a) How many integers from 1 through 1000 are multiples of 2 or multiples of 9?
Let A = the numbers from 1 through 1000 that are multiples of 2

Let B = the numbers from 1 through 1000 that are multiples of 9

|A| = 500 |B| = 111 (they go from 9 = 1*9 to 111 = 111*9)

A ∩ B is the number of numbers which are multiples of 18. There are

55 of these (from 18 = 1 * 18 to 990 = 55 *18)
So |A U B| = |A| + |B| - | A ∩ B | = 500 + 111 – 55 = 556

b) How many integers from 1 through 1000 are not multiples 2 or 9?
1000 – 556 = 444
c) What is the probability that an integer chosen at random from 1 through 1000

is a multiple of 2 or 9?
P(multiple of 2 or 9) = 556/1000 = 0.556
d) What is the probability an integer chosen at random from 1 through 1000

is not a multiple of 2 or 9?
P(not a multiple of 2 or 9 ) = 1 – 0.556 = 0.444

or

P(not a multiple of 2 or 9 ) = 444/1000 = 0.444
#4. A Poker table contains 15 seats. What is the maximum number of people

who can sit down if they wish to play 5-card Poker from a standard 52 card deck?

52/5 ┐ = 10 (52 “pigeons” to go into 5 “pigeonholes”)


#5. Let R be the relation on the set {0,1,2 } defined by R = {(0,1), (1,2) }
a) What is R U R-1
R-1 = {(1,0), (2,1) }
So R U R-1 = { (0,1), (1,2) (1,0), (2,1) }
b) What is the reflexive-closure (R)


Reflexive-closure (R)= {(0,1), (1,2) (0,0), (1,1), (2,2)}

c) What is the symmetric-closure (R)
symmetric-closure (R) = { (0,1), (1,2) (1,0), (2,1) } (which you may notice is R U R-1 )
d) What is the transitive closure(R)
Transitive-closure (R) = {(0,1), (1,2) (0,2)}
e) What is R o R?
R o R = {(0,2)}

#6. Let R be a relation over the integers defined by m R n  4 | (m –n);

that is m ≡n mod 4


  1. Show that R is an equivalence relation


reflexive because 4 | m – m

symmetric because if 4 | (m-n), then 4 | (n – m)

symmetric because if 4 | (m – n) and 4| (n – p) then 4 | (m – n) + (n – p) or 4 | (m – p)



  1. Show the equivalence classes for R (that is, show the members of each equivalence class)


[0] = {…, -4, 0, 4, 8, …}

[1] = {…, -3, 1, 5, 9, …}

[2] = {…, -2, 2, 6, 10, … }

[3] = {…, -1, 3, 7, 11, …}
#7. Ninety-one students took exam #1. Could each of these 91 be friends with exactly 3 others in the class? Why or why not?
No possible because if we try to draw an acquaintanceship graph, each of the 91 vertices would have degree 3 for a total of 273. But the sum of the degrees of a simple graph must be even.
#8. For the following, draw the simple graph or state why none exists using rules of graphs:


  1. 5 vertices of degrees 0, 1, 2, 3, 4


Not possible because all simple graphs with #vertices > 2 must have at least two vertices of the same degree


  1. 5 vertices of degrees 2, 2, 2, 2, 2


A pentagon will do nicely


  1. 5 vertices of degrees 1, 2, 2, 3, 3

Not possible because the sums of the degrees must be even (also there must be an even number of vertices of odd degree)


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