### 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?

1. My questions:

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

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

2. 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: 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:
“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
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

i’m looking for
$4\binom{13}{5}$
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?

3. 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.

4. 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…

5. vlorbik

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…

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

Covering my error with a cute exit.

7. vlorbik

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…
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
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
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,
like this (using the notations of
the exam-prep materials;
$\null_{12} P_4$ 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.

• 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.

8. 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).

9. 40!/[10!10!10!10!]
(shorter but harder to type;
a comment on my earlier comment).

10. 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
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
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
${4\choose 0}+ {4\choose 1} + {4\choose2}$
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.

11. 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?

${52 \choose 2} = (52*51)/(2*1) = 1326$

$4*{4\choose 2} = 24$

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.)

12. 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.