### from file folder COMBO.131 (“finite math” problems)

Exam 2. August 23, 1999

1.

A club with 12 members elects a President, a Vice-President, a Deputy Vice-President, and a Go-fer (no member can serve in two different offices). How many different election results are possible?

2.

A student believes that one-fourth of the 40 multiple-choice problems on a certain exam are to be answered “A”; one-fourth are to be answered “B”; one-fourth “C”; and one-fourth “D”. Give a formula for the number of possible patters of answers meeting these conditions. Hint: consider the 40-letter “word”

AAAAAAAAAABBBBBBBBBBCCCCCCCCCCDDDDDDDDDD

3.

An “outfit” for the day consists of a choice of T-shirt, a pair of jeans, and a pair of shoes. I have 30 T-shirts, 4 pairs of jeans, and 2 pairs of shoes to choose from. How many different “outfits” are possible?

4.

Write down the sample space (16 outcomes) for the experiment “toss a coin four times”.

a.

What is the probability that **exactly** two heads occur?

b.

What is the probability that **at least** two heads occur?

5.

I have 6 math books, 3 novels, and 1 *Bible* on my coffeetable. I decide to put four of these books in my backpack.

a.

How many different sets of four books do I have to choose from?

b.

How many of these do *not* include the *Bible*?

c.

How many include *exactly one* novel?

6.

A “Texas Hold’em” hand is a set of two cards chosen from a standard deck. How many different Hold’em hands are there?

How many Hold’em hands are “high pairs”—defined here as a pair of Jacks, Queens, Kings, or Aces?

7.

How many distinguishable arrangements are there for the letters of “ELEEMOSYNARY”? (Bonus Point: what does this word mean?)

8.

A billionaire awards large cash prizes to 5 lucky banjo players. On Monday, he will give away 1 million dollars; on Tuesday, 2 million; and so on until Friday, when he will give away 5 million. There are 11 banjo players on his short-list. In how many ways can the prizes be awarded?

Quiz. February 21, 2002

1.

For the sample space S={h,a,d,n,o,e,x,i,t,s}

and events E = {d,e,a,t,h} and F={t,a,x,e,s} determine the indicated events

a. E’

b. F’

c. EuF

d. (EuF)’

e. E’uF’

f. E&F

g. E&F’

h. (E&F)’

2.

Suppose P(E)=0.3, P(F)=0.4, and P(E&F)=0.2. Find **a.** P(E’) and **b.** P(EuF).

3.

A couple has four children. Write out all (sixteen) of the possible birth orders: BBBB (four boys), BBBG,…,GGGG. What is the probability that there are more girls than boys?

4.

A poker hand is a set of five cards chosen randomly from the standard deck. A flush is five cards all of the same suit. Find the probability of a flush (hint: four times the probability of a *diamond* flush).

5.

How many distinguishable arrangements are there for the letters of PARAMETER?

6.

How many (seven digit) phone numbers can be created *without* using a “0”?

How many of them have no *repeated* digit?

7.

A gunsmith has 4 locks, 5 stocks, and 6 barrels. He selects one of each. In how many ways can he make his selection?

March 6, 2010 at 2:20 pm

My questions:

Question 5 and (we can distinguish 2 novels, no?)

Quiz Question 4 (is a straight-flush a flush?)

March 7, 2010 at 1:15 pm

5. we can, yes.

thus

*********************************************

S = {

m_1,m_2,m_3,m_4,m_5,m_6,

n_1,n_2,n_3,

B

}

is given.

how many (of the 2^10 = 1K)

subsets (X) of S

have the property that

a.) X has four “points”.

\*comment: in our offices with the doors

\*comment: closed, graduates of math

\*comment: departments refer to “points”

\*comment: of a “space” more or less

\*comment: interchangeably with “elements”

\*comment: of a “set”

#X = 4

in other words.

(a handy abbreviation.)

b.) #X=4 and B\not\in X

c.) #X=4 and X\intersect {n_1, n_2, n_3}\not\equal\nullset

********************************************

is an elaborate rephrasing of 5.)

on the much more subtle question

“is a straight flush a flush”.

first of all: good question.

now.

my immediate reaction:

i read the problem again.

“a flush is five cards of

the same suit.”

i imagine myself in the position

of a reader knowing nothing of

the rules of poker…

but knowing the suits and ranks

of the 4 by 13 array beloved

of all teachers of probability…

and i take this definition as i

more or less *must*:

inclusive of *all*

sets-of-five having

no second suit.

there’s a great deal hidden here.

the teacher hopes the student

knows the deck… we’ve practiced

some in class and whatnot.

the student hopes the teacher

isn’t unconsciously *loading* the

deck (again) with culturally-

-biased assumptions of which

they may only be very dimly

aware.

“extra credit”… very likely just

a minute of fuss made in class by a

widely grinning instructor…

for a student just for

(coherently) asking this question.

i’m looking for

in other words.

considering straight-flushes

as *excluded* is a harder

question… of the type i’d

never have dared put on

a quiz for classes like “131”.

maybe as a take-home.

is a royal flush a straight flush?

does the buddha have a cat nature?

March 7, 2010 at 6:07 pm

As always, when I introduce cards, I slowly run through suits, colors, ranks, face vs number, etc, etc. “You probably know all of this, but be patient, I am certain that there are 2 or 3 of you for whom this is not familiar information” (2 or 3, not 1 or 2, because if 1, and you don’t know, then I’ve singled you out. With 2 or 3, if you don’t know, you have a secret compadre).

Anyhow, when the kiddies write their course evaluations (I have teased out fairly honest, fairly thoughtful, short answer type questions), this year I got a complaint about not teaching the cards well enough for those who don’t know them well. Hmmmph. I need to be more careful.

For the non-player, all flushes are flushes. For the player, straight-flushes are not flushes, and royal flushes are simply the highest ranked straight-flushes.

March 8, 2010 at 5:25 am

A student, a player, and a math major, thinks I’m wrong (he didn’t say it like THAT, in fact, I posed the question roundabout), but it’s enough for me to want to back off the flushes…

March 8, 2010 at 11:38 am

we like to use language like

“any straight flush beats any flush”…

but in a math-class setting it’s best

in my opinion to regard this as *slang*.

more evidence not that any is needed

that “real life” problems are *harder*

than “abstract” problems.

“real life” definitions are *always* messy

(because nothing *is* what it *seems*).

*only* in the realm of the purely symbolic

can we ever actually mean *exactly* what

we say…

March 8, 2010 at 12:51 pm

Indeed. A pair of Jacks is better than nothing, and nothing beats a straight-flush…

Covering my error with a cute exit.

March 12, 2010 at 6:27 pm

answers and remarks

1)

P(12,4)=12.11.10.9=11880

the middle one is hugely

to be preferred if one is

allowed only *one*…

and is not provided a context.

which is absurd i hope but bear

with me for purposes of argument.

because nobody ever wants to believe

contexts matter or something and if

i can’t put some brief stop to it in a

comment on my own post in my own

blog well i’m in even more trouble

than i think i am. (which, as you can

imagine, gets *tiring* when it keeps

on *happening* over and over.)

the middle one, then,

12.11.10.9,

sort of *tells its own story*.

it tells the story much more clearly

to well-prepared minds, of course,

like any story… on parle francais ici?

me neither, but you get the idea. so.

what’s this *preparation*?

in my work, nominally, it would consist,

minimally, of having seen a lecturer

go through at least one example

a great deal *like* it.

the “big picture” one hopes to have

instilled with this work consists,

for problems like this, of the idea

that one has here a *sequence*

of things we can think of as

occurring in over *time*;

that one can then divide-and-conquer

by creating a “model”

(which word need not and indeed generally

ought not escape the lecturer’s mouth):

*first* we’ll pick the president (the choices are 12),

*second*, a vice (11)…

the physical manifestation of our model,

i now make it a point to remark, might

very well look like

_ _ _ _

: four spots on the board

signifying, first, hmmm.

well, this is where it gets interesting

because you’ve got to keep trying to

pop people’s expectation-balloons…

which *will* keep getting in the way…

while still radiating confidence that

once it all makes sense, it’ll be,

and i make a point never to write this

on the board without quote marks,

“easy”.

once you know even a *little*

of what’s going on, you’ll know

those _ _ _ _ spots are gonna have

*numbers* in ’em… but!

during the lecture, the teacher… “they”, them right there

without the gender… is gonna point at the spots

on the board and say things like

“when we’re right here, we’ve

already picked a president

and a vice, and so…”

(blah, blah, blah… writes “10”).

and they’re gonna do well to do so;

i’d do it myself and have many times.

and maybe they’ll even write

P V D G

under the S.O.B.’s…

i’ll have done something like this

many times too

or maybe the old

1st 2nd 3rd 4th

or… you might as well,

there’s no telling *what*’s

gonna get in their notebooks anyhow

1 2 3 4

.

now, to the *unprepared* mind

_ _ _ _

1 2 3 4

and

12 11 10 9

__ __ __ __

look so much alike that the wonder

is that anybody gets through any

of this alive.

because, while, believe it or not,

*i* can satisfy pretty much any

reasonably patient speaker of english

that they’ve understood what’s going

on in this problem, with very little fuss,

and be right, there aren’t very *many* people

who’d make this (elaborate) claim and most

of those who would are dead wrong.

and once you put those scribbles

on the board *and* start talking about ’em,

there’s this double discourse.

and the level of formality is always

*changing* and is very often

*contested*.

nobody wants to talk about any of this

and for good reasons. we just want to

talk about solving *math* problems

for hecksake; we’re math *majors*.

anyhow, 11880. get a grip man.

as to P(12,4).

when you (okay, now *you’re* the gender-free

teacher! see how *you* like it!) work with

students who are preparing to be

*tested* on this kinda thing,

and they’re *ready*, they’ll

comfortably rattle off an answer

like this (using the notations of

the exam-prep materials;

maybe).

this will *not* suffice for

all examiners, of course,

since

P(12,4) = 12!/[4!(12-4)!]

may be about as far as

certain quit-trying-to-explain-its

will ever get in “simplifying”

this result without a calculator

(and somebody to punch its

buttons for them; you will

be blamed for this).

2)

forty-bang over product tenbang,tenbang,tenbang,tenbang.

i’m very tired. this is harder than it looks.

March 12, 2010 at 9:45 pm

i’m not sure what your point is, but #2 is 40 flags, 10 of each color, how many ways can they be hung in a line? if you’re talking examples.

what i got out of this is – when we try to explain these things the students have to hold different levels of abstraction in their heads at once, and that’s part of what makes it hard to get, until you get it.

March 13, 2010 at 2:18 pm

right: the flags-on-a-pole problem is

another version of this situation.

there are

10 Aqua(lung)

10 Blue (\”oyster c\”ult)

10 Creem-colored

and

10 Deep purple

flags etcetera (AAA…DD).

the “letters” version has the virtue…

or misfeature… “one less layer

of abstraction”: the objects we

“move around” on the page

actually *are* letters.

(we might use colored pencils

or chalk for eyecandy vividness

for purposes of presentation

and even make it our life work

but we won’t be doing math but

page design or animation or

selling selling selling all the time).

March 13, 2010 at 2:21 pm

40!/[10!10!10!10!]

(shorter but harder to type;

a comment on my earlier comment).

January 19, 2011 at 2:12 pm

3.

An “outfit” for the day consists of a choice of T-shirt, a pair of jeans, and a pair of shoes. I have 30 T-shirts, 4 pairs of jeans, and 2 pairs of shoes to choose from. How many different “outfits” are possible?

30*4*2 of course:

the “multiplication principle”

at (almost) its simplest.

in particular, the situation is simpler than

that of problem 1. — where in some sense,

the “choices” (but not the *number* or choices)

associated with the various spots of the

__ __ __

diagram are *not* independent

(a person chosen for president can’t

be *also* chosen for vice-president,

but the choice of T-shirt does *not*

affect the choice of jeans.

one introduces the matter of “independence”

at one’s own peril here of course…

my practice was to talk about it

a little bit *without* “reading it

into the notes” (by writing about

it on the board). we want…

very much… for the students to

see this as a very simple situation.

it’s more or less standard to present situations like this

with so-called “tree diagrams”. this shouldn’t be overlooked

altogether but isn’t the be-all-end-all that you’d be led

to expect from some presentations: just as with the

fill-in-blanks notation i was fussing about upthread,

there are certain “hidden assumptions” about filling

these diagrams in… and making ’em explicit isn’t

*necessarily* the direction of “more clarity”.

remember: students don’t like to read (and

*particularly* don’t like to read *unfamiliar material*.

4.

Write down the sample space (16 outcomes) for the experiment “toss a coin four times”.

a.

What is the probability that exactly two heads occur?

b.

What is the probability that at least two heads occur?

hhhh

hhht

hhth

hhtt

hthh

htht

htth

httt

thhh

thht

thth

thtt

tthh

ttht

ttth

tttt

a) #{hhtt, htht, htth, thht, thth, tthh} = 6

(read ’em off the list!)

so the probability is 6/16 (= 3/8)

(full credit for .375 of course though

this is to be discouraged… decimals

don’t always “work out exactly”…)

b) read ’em off the list again: 11/16.

it may (or may not) be worth mentioning

in a given context that one has

in the numerator here.

writing out the list itself?

i’ve long lost track of the number

of classes i’ve “drilled” this into.

for the “just show me what to do” crowd:

compute the number of “outcomes”: 2^4 =16.

then write out (downwards) h, t, h, t, …

until you’ve got 16 lines.

go back to the top line and add (at the front

if you want my exact display) a letter to each

line; follow the pattern h, h, t, t, h, h, …

(“counting by twos”). then count by 4’s; then 8’s.

done.

January 19, 2011 at 5:51 pm

5.

(see the first comment in this long thread.)

6.

A “Texas Hold’em” hand is a set of two cards chosen from a standard deck. How many different Hold’em hands are there?

How many Hold’em hands are “high pairs”—defined here as a pair of Jacks, Queens, Kings, or Aces?

7.

How many distinguishable arrangements are there for the letters of “ELEEMOSYNARY”? (Bonus Point: what does this word mean?)

12!/(3!*2!) = 39 916 800

“eleemosynary” means (something like) “charitable”.

(nobody knew this on the day if i recall correctly.)

January 19, 2011 at 5:55 pm

8.

A billionaire awards large cash prizes to 5 lucky banjo players. On Monday, he will give away 1 million dollars; on Tuesday, 2 million; and so on until Friday, when he will give away 5 million. There are 11 banjo players on his short-list. In how many ways can the prizes be awarded?

11*10*9*8*7

(from first principles) or

P(11, 5)

(using the “permutation formula”).

the dollar amounts are of course there

to make “order matters” vivid.

everybody’s always thinking about money.

August 22, 2014 at 6:44 pm

http://samjshah.com/2009/04/14/composition-of-functions-and-their-inverses/

sam shah on the cosmic yin-yang.

with no particular place to go.