Intro Prob Hw1

Liv Danielson
February 2023

1 Problem 1
1. How many sequences of length 3 can be formed from the digits 1,2 and 3
if repetition is allowed?
The length is 3 and there are three spaces to put the three digits, 1,2,
and 3. Looking at the first space, there are three choices. For the second
space, there are also three choices as repeats can happen and there are
only three digits to choose from. Finally, there are three choices for the
final space. Multiplying the choices together you get 3 * 3 * 3. So the
answer is
33
2. How many sequences of length 3 can be formed from the digits 1,2 and 3
if repetition is not allowed?
Part 2 follows part 1 very closely except for the phrase ”repetition is not
allowed”. Again, there are three spots for the three digits to go into,
except this time instead of three choices each time, for the first spot there
are three choices, then the second spot, there are two choices, finally, for
the last spot, there is only 1 choice. So the answer is 3 * 2 *1 or

3!
3. How many sequences of length 10 can be formed from the digits 1,2 and 3
if repetition is allowed?
With the sequence having a length of 10, and that repetition is allowed,
each of the spots in the sequence has 3 choices out of the three digits
available. Therefore the answer would be 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3
* 3 or
310
4. How many sequences of length 10 can be formed from the digits 1,2 and 3
if repetition is not allowed?
Because the question is asking for a sequence of length 10, and there
only three digits, and repetition is not allowed, the highest length of the
sequence that can be made is 3, therefore it cannot happen.

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 0 or not possible

5. How many sequences of length 10 can be formed from 0,1,2,3,4,5,6,7,8 and

9 if repetition is allowed?
Looking at the number of spaces, because the question asks for sequence
and not numbers, one can have 0 be at the beginning of the sequence,
where it couldn’t be for numbers. Therefore there is 10 choices for each
spot in the sequence resulting in

1010

6. How many sequences of length 10 can be formed from 0,1,2,3,4,5,6,7,8 and

9 if repetition is not allowed?
As there are 10 spots in the sequences because it is of length 10, and there
are 10 digits, the amount of choices for the first spot is 10, then for the
second spot is 9, and so on and so on until you reach the end. Thus the
answer being 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 or

10!

2 Problem 2
A sandwich is made by choosing one piece of flatbread, one slice of ham, and one
slice of cheese where the choices drawn equally likely at random from four pieces
of flatbread, from three slices of ham, and from three slices of cheese. We later
learn that one piece of flatbread, one slice of ham, and one slice of cheese were
spoiled. Compute the probability the sandwich made was good, i.e., it will not
contain any spoiled items. What’s the probability that the sandwich contained
at least one bad item?
To start with there are 4 choices of flat breads, 3 choices of slices of ham, and
3 choices of slices of cheese. But when there is one for each which are spoiled,
then the choices drop down to 3,2,2 respectively. Looking at the first part of
the question, the total choices possible are 4 * 3 * 3 which is 36. Thus, 36 is |Ω|.
For the Top, you look at the number of choices for each spot. For the choices of
good flat breads, there are 3, then 2 for the ham and 2 for the slices of cheese.
Therefore the answer to the first part of the question is 12/36 or
1
3

For the second part of the question, we can use the complementary rule
because it is asking the complement of the other part of the question which is
the first part was asking for the probability of the sandwich has good ingredients
and the complement to that would be the probability the sandwich has a bad
ingredient. Using the complement rule, then it is exactly 1 - 1/3 so the answer
to the second part would be
2
3

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3 Problem 3
Each of three people toss a coin. What is the probability of someone being the
odd man out? This means that two of them obtain an identical outcome, while
the odd man gets a different one.
First you can write out all the possibilities of the outcomes when the three people
toss a coin. Those would be HHH, HHT, HTH, HTT, TTT, TTH, THT, THH.
In total, there are eight possible outcomes. Six of these outcomes, satisfies what
the question is asking, which is when someone is the odd man out, or that there
are two of either heads or tails and then one of the other. These six outcomes
would be HHT, HTH, HTT, TTH, THT, and THH. Therefore the answer is
6 3
8 or 4 or .75

4 Problem 4
An urn contains 3 green and 4 yellow balls. We draw three (3) balls one by one
without replacement (such that at each draw every ball is equally likely to be
selected).
1. Find the probability that the colors we see in order are green, yellow, green.
Starting off, there are 7 balls in total. The sequence which is needed for
this problem is G Y G. The total amount of choices are 7 then 6 then 5, as
it only needs to be a length of 3 and since there isn’t replacement, every
time a ball is drawn, the total amount of balls which can be drawn from
go down by 1. Then looking at the certain choices, there are 3 choices for
the first spot as there are 3 green balls. Then there are 4 choices for the
second as there are 4 yellow balls. Finally there are 2 choices for the third
as there is now 2 green balls since one was chosen for the first spot. In
other words, for the first spot, there are 3 choices out of 7 balls, then for
the second there is 4 choices out of 6 balls, then 2 choices out of 5 balls.
So the answer is
3 4 2
7 * 6 * 5

2. Find the probability that our sample of 3 balls contains 2 greens and 1
yellow.
This is very similar to section 1 of the problem. Any two greens and 1
yellow could be GGY, GYG, or YGG. The first part of the question tackles
the scenario of GYG. One way to answer this question is by adding up
the probability of the other two scenarios as well, another way would be
to multiply one of the scenarios by 3 because it comes out to the same
answer. With GGY, it would be 37 * 26 * 45 and then for YGG, it would
be 74 * 36 * 25 . Adding all of these up, one would get ( 73 * 26 * 45 ) + ( 47 * 36
* 25 ) + ( 73 * 46 * 52 ). This is one version of the answer, however, because
each scenario is just the same as the other one just in a different order,

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 so one could multiple any one of the scenarios by three, as there are three
different scenarios which could answer the question.

( 37 * 2
6 * 45 ) + ( 74 * 3
6 * 52 ) + ( 37 * 4
6 * 25 ) or 3∗( 37 * 4
6 * 52 )

5 Problem 5
Mark goes out to dinner twice a week. If he chooses the days to go out at
random, what is the probability that

he goes out on exactly one weeknight?
First, trying to find |Ω|, it is 72 because in total, there are seven days of the
week. Of those seven days, he wants to go out twice, so there are seven choose
two ways he can choose to go out. For the top, since he only wants one of
the
days that he goes out to be a weeknight, for the first choice, there are 51 as
there are five options of weekdays to
go out and he only needs to choose one of
them. For the second spot, it is 21 as there are only two weekends to choose
from. It is the same if the first spot was for the weekend day and the second
spot was for the weekday. Therefore the answer is:

((51)∗(21))
(72)

6 Problem 6
A bus starts with 5 people and stops at 7 different stops. Assume that passengers
are equally likely to depart at any stop, no new passengers enter the bus, and
all passengers eventually leave the bus.

1. In how many ways can these 5 people leave the bus?

Because no new passengers get on the bus, and everyone will get off the
bus. There are five spots and each one has 7 choices because for each
person, there are seven choices for each 5 person as two people could get
off at a stop and all are equally likely to depart at any stop. So the answer
is:

75

2. Find the probability that no two passengers leave at the same bus stop.
Using the previous question to help answer this one. The previous section
of problem 6 is the |Ω| or the denominator. So 75 = |Ω|. Then because no
passengers will leave at the same stop, the first person has 7 choices, then
6 choices, then 5 choices, then 4 choices, then 3 choices. This is called
seven falling factorial five.

7∗6∗5∗4∗3
75

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 3. Find the probability at two (or more) passengers get off at the same stop.
For this problem, it is the complement to section 2 of this problem because
it is the opposite of no two people leaving at the same time, this one is
the probability two or more passengers get off at the same stop. Therefore
the answer is

1- ( 7∗6∗5∗4∗3
75 )

7 Problem 7
We plan to turn over the first 3 cards in a well-shuffled deck of 52 cards. Com-
pute the probability that both the second and third cards are the same suit.
There are two cases where the problem is true. One case is when all the cards
are the same suit,and the second is when only the second and third cards are the
same case. For both of the cases, the denominator or |Ω| is equal to 52 ∗ 51 ∗ 50
because for each of the three cards which are turned over, there are 52 cards,
then 51, and then finally 50. For the first case, first one needs to choose the
suits, which there are four suits and only one would
need to be chosen, as all
the cards would be the same suits. That would be 41 . Then because there are
13 cards in each suit, for the first choice there are 13 then 12



1 1 and finally
11


1 . Then you need to add the second scenario in. For this, there can be two
suits as the first card can be a different suit from the second and third. The
denominator is the same, however the
top is a little different. Because there
are now two choices for suits, it is 42 . Then for the first card, because it is a
different suit, it is 13 13



1 , as well as for the second
spot is also 1 as it is from a
different suit. Finally, for the last spot, it is 12
1 . The answer is

((41)∗(13 12 11
1 )( 1 )∗( 1 )) ((42)∗(13 13 12
1 )( 1 )∗( 1 ))
52∗51∗50 + 52∗51∗50

8 Problem 8
There are 4 boys and 5 girls to seat in a row of chairs.

1. How many seating arrangements are possible?

In total, there are nine spots for each of the people to go into as there are
nine people in total. For the first spot, there are nine choices, then eight,
then seven, then six, and all the way down to one. This is also calling 9
factorial.

9!

2. How many seating arrangements are possible if the children must alternate
gender?

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 Because there are more girls than boys, a girl will have to be seated first
as then the children will alternate. Therefore there are five choices for the
first seat as there are five girls. Then there are four choices for the second
seat as there are four boys. Then four choices for the next spot as there
are now four girls. This continues until all the children are seated. As this
is happening, it looks like 5 * 4 * 4 *3 * 3 * 2 * 2 * 1 * 1, this is also equal
to

5! ∗ 4!

3. How many seating arrangements are possible if Fred (a boy) and Carrie
(a girl) must be seated next to each other?
First, you can combine Fred and Carrie into one entity, lets call F. Now
there are eight total people to sort, as it now is three boys, 4 girls, and one
F. So to sort eight people, that is eight factorial, but then you also need
to multiply how to sort what is in the big entity that was made, which
is 2 factorial or just 2 because there are only two things in the entity.
Therefore the answer is

2! ∗ 8!

4. How many seating arrangements all the boys seated next to each other?
This can be answered using the same thought as in 8.3. Combining all
the boys together we can say that it equals B. Therefore, there are now
5 girls and 1 B, so there are six total things to sort. Sorting six things
without repetition is 6 factorial. Then you still need to sort whats inside
the capital b. There are four things inside B, so being able to sort four
things is four factorial, as there is four choices for the first spot, then
three then two then one. So just like in 8.3, the answer to this part of the
problem is

6! ∗ 4!

5. How many seating arrangements all the girls seated next to each other?
This is very similar to 8.4, except instead of lumping all the boys together
into B, you are lumping all the girls together in a G. Therefore there is
now six total people to sort, as there is 1 G and 5 boys, which equals six
in total. So for sorting five total people, there are 5 factorial ways to sort
them. Then looking at whats in G, there is also five people, which means
there is also 5 factorial ways of placing them. Thus the answer is

5! ∗ 5!

6. How many seating arrangements all children of the same gender seated
next to each other?
Putting all the boys in one category and all the girls in another category,

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 there are two possible ways to sort these two categories, as one of them
the boys can go first and the second one the girls can go first. Therefore
there are 2 factorial ways to sort B and G. Then within each of them, for
the B, there are 4 boys within which allows for four factorial ways to sort
them. Then with the girls, there are 5 girls in G, so there are five factorial
ways to sort them. So the answer is 2! * 4! * 5!

2! ∗ 5! ∗ 4!

7. Re-do 8.1 when there are 5 boys and 5 girls.

Looking back at how I did 8.1, you take the total amount, which in this
case is now 10, so there are 10 spots to spot 10 people into, which means
that there are 10 choices for the first spot, then nine for the second, then
eight for the third, and all the way down until 1 choice for the last spot.
So the answer is 10 factorial.

10!

8. Re-do 8.2 when there are 5 boys and 5 girls.

This is very similar to 8.2, except now there are an even number of boys
and girls. This means that there is no one gender which must be seated
first. So you need to mulitple my 2 or 2 factorial. Because each group has
five in each, each has five factorial wasy of sorting them. So the answer is
2! (5! * 5!)

2! ∗ (5! ∗ 5!)

9 Problem 9
1. The letters in the word DISPARE are rearranged at random so that every
possible anagram is equally likely. What is the probability that it still spells
DISPARE?
First one needs to find |Ω|, which would be the total amount of choices
with the total number of letters. In DISPARE, there are 7 letters and
none of the letters repeat so, the way that you can sort them is 7 factorial
ways. As it is seven choices, then six, and so on. Then to figure out the
numerator, there is one 1 possible way for the letters in DISPARE to make
DISPARE. Thus the numerator is 1.
1
7!

2. The letters in the word MIAMI are rearranged at random so that every
possible anagram is equally likely. What is the probability that it still spells
MIAMI?
This is similar to the first part, except there are repeated letters. So

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 finding |Ω| is a little different. There are two sets of two letters which
repeat, so each can be sorted two factor ail ways, and the letter by itself,
A, can be sorted one factorial ways or just one. So |Ω| is 2! * 2! * 1!.
Then for the numerator because there are in total five letters and there is
only one way which they all spell MIAMI, the numerator is 1/ 5!. So the
answer is
1
( 5! ) 2!∗2!
2!∗2!∗1! or all this simplified is 5!

10 Problem 10
Consider the word BOOLAHUBBOO.
1. How many anagrams are possible?
Finding |Ω| is the same as in 9.2 because there are repeating letters. O is
repeated four times, B is repeated three times and the rest are repeated
never. This means that |Ω| is 4! * 3! * 1! * 1! * 1! 1!. For the four
O’s there are 4 factorial ways to place them and for the three B’s there
are three factorial ways to place them. The rest of the 1! is just equal
to 1 so it doesn’t matter if they are included or not. For the numerator,
it is just the total amount, discounting repeating letters, so counting up
all the letters, there are 11 in BOOLAHUBBOO, so the numerator is 11
factorial since there are 11 factorial ways to place 11 letters.

11!
4!∗3!

2. How many of these anagrams end BOOBOO?

Putting BOOBOO into one big category, called Z, means that there are
now 5 letters in total because there is an L, A ,H ,U and a Z. with 5 letters,
you can sort them 5! ways. Because they will all end the same with Z or
BOOBOO, you do not need to do anything within Z, just with the rest of
the letters, so the answer is 5 factorial.

5!

3. How many anagrams have all the B’s grouped together?

Let’s start like the others by grouping all the B’s into a category called
P, so now to find |Ω|, it is similar to part 1 of the problem, except this
time there is only one repeated letter, which are the four O’s. Therefore
the denominator will simplify to just 4 factorial because there are 5, 1
factorials which all simplify to 1. For the numerator, you need to take
into account how many total letters there are, but not counting all the B’s
only the one category P. So there are 9 letters total, which can be placed
9 factorial ways.

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 9!
4!

4. How many anagrams have all the B’s grouped together and all the vowels
grouped together?
One can group all the B’s into one category labeled P and all the vowels
in another category labeled V. So first, in total, there are 4 letter,s P,
which represents all the B’s, V, which represents all the vowels, L and H.
These can be grouped 4 factorial times. Then you need to group all the
B’s together, however it will just be 3! over 3! because in total, there are
3 letters, but the only letters in it are also repeated three times, so that
will be equal to 1. Then looking at V, the total number of letters are 6,
which can be grouped 6!, however, the only repeating letter is O, which
is repeated 4 times, which is 4! times to be grouped. The other vowels
would be 1! which equates to 1. So the answer would look like

4! ∗ ( 3!
3! ∗
6!
4!) which can simplify to 6!

11 Problem 11
I have 3 one dollar bills, 2 five dollar bills and 5 ten dollar bills. How many
ways can I distribute these 10 bills to 10 children so that each child gets one
bill? Treat the bills of equal value as indistinguishable please.
Figuring out |Ω|, is very similar to when there are repeating letters, but in
this case, there are repeating bills. There are 5, one dollar bills, which can be
arranged 5 factorial ways. There are 2 five dollar bills which can be arranged
2 factorial ways. Finally, there are 3 one dollar bills, which can be arranged 3
factorial ways. so |Ω| is equal to 5! * 3! * 2!. To find the numerator, one counts
up the total amount of bills, which is 10. Those can be arranged 10 factorial as
no child can get two bills.
10!
5!∗3!∗2!

12 Problem 12
We turn over the cards of a well-shuffled standard deck of 52, and we observe
the sequence.
1. How many sequences have all cards of the same suit together?
First, there are 4 suits in a deck of cards, each with 13 cards in each suit.
So the total number to arrange 4 suits, would be 4 factorial. Then within
each suit, there are 13 cards, which can be chosen 13! ways because there
are no repeats, and that have to be multiplied out 4 times because there
are 4 suits. So the answer is 4! * (13! * 13! * 13! * 13!)

4! ∗ (13!4 )

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 2. How many sequences have all the clubs together?
Looking at all the clubs together as one, we can put it into one category
called C. Then the total amount of cards is 39 + C, which is equal to
40 cards in total, which, because there are no repeats, comes out to be
arranged 40 factorial ways. Then you need to take care of within the
category C. Within it, there are 13 cards, which because there are still not
repeats, can be arranged 13 factorial ways. SO the answer is 40! * 13!

40! ∗ 13!

13 Problem 13
There are 40 flavors of Insomnia Cookies. If Harry plans to make a box of 5
cookies of 5 different flavors, how many different boxes are possible?
In total, there are 40 flavors, of those flavors, you need to choose 5 of those
flavors, 5 times over. So the total amount is 40, then one is choosing 5, so the
answer is
40


5

14 Problem 14
Kevin works at the Applied Physics Laboratory. There are 21 projects being
worked on, and he must choose 4 to supervise. In how many ways can Kevin do
this?
Looking at the question, there are 21 total projects to choose from and Kevin
needs to choose 4, so the answer is 21 choose 4.
21


4

15 Problem 15
Pn Pm
Show that k=1 (k nk ) = n(2n−1 ) by using the fact that 2m = j=0 m



j for
positive integer m
First we
Pm can say
that supposed 2 =n x + 1 or 1 + x. Then you plug that into
m
2m = j=0 j and you get 1 + x =, which is the binomial theory so you
can expand that out. (Also it doesn’t matter if you use m or n because they
are just placeholders so one canPchoose any

variable to be a placeholder.) Ex-
n
panding 1 + xn , it is equal to k=0 ( nk xk (1n−k )). The 1n−k will always be
equal to 1 as 1 to any
Pnpower
is always 1, so it doesn’t have to always be written
down. Then from k=0 ( nk xk (1n−k
)), taking the derivative of each side, will
Pn
result in n(1 + xn−1 ) =
k=0 ( nk k(xk−1 )). The way that this equation can
Pn
now look like k=1 (k nk ) = n(2n−1 ), if by replacing x as 1 since 2 = 1 + x,
Pn n

k−1
which makes x = 1. Therefore, you get n(2n−1 ) = (
k=0 k k(1 )). The

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1k−1 will also always equal to 1 because 1 to any power is 1. Now it is al-
most like the original of what
Pn wen
are k−1trying to prove. When Pnk = 0,
and0−1
it is
n
plugged into n(2n−1 ) = (
k=0 k k(1 )), it is equal to (
k=0 0 0(1 )).
The most important thing to know is that everything in the summation is
that it is multiplied
Pn
by k P
so when
k = 0,
the summation is multiplied
by 0.
n Pn
Therefore
k=0 (
nx x) = k=1 ( nk
k) + n0 ∗ 0. Thus, 2m = k=0 ( nx x) =
Pn n n Pn n
k=1 ( k k) + 0 ∗ 0 = k=1 (k k )
Pn n

Pn n

n

Pn n


2m = k=0 ( x x) = k=1 ( k k) + 0 ∗0= k=1 (k k )

16 Problem 16
The Non-cents Corporation has 4 entry-level job openings and 37 female and
7 male apply for these openings. Assume the applicants are all equally quali-
fied and the decision to hire will be done so that each subset of 4 from the 44
candidates is equally likely to be hired. Compute the probability that 1 or fewer
females are hired.
Dealing with the denominator first, there are in total 37 + 7 people that can
be hired into 4 jobs, which is
44 people with 4 positions to try to get hired for.
Thus the denominator is 44 4 . For the numerator it is a little harder. One needs
to take into account the two separate scenarios. The first one is if none of the
females get hired. Then the only total population of people which the 4 jobs
will be chosen out of is the 7 males. Thus the numerator for scenario 1 is 74 .

For the second scenario, one female is hired, along with the rest being males.
Therefore, the first position would be 37


1 as there are a total of 37 females, but
only 1 would get a job. Then that would
get multiplied to the rest of the ways
to choose the other three jobs which is 73 . The final answer would look like

(74)+((37 7
1 )(3))
44
(4)

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