Download 50 Counting Questions - Solutions

50 Counting Questions - Solutions
Prob-Stats (Math 3350) Fall 2012
Formulas and Notation
n!
• Permutations: P (n, k) = (n−k)!
, the number of ordered ways to permute n objects into
k bins.
n!
, the number of unordered ways to arrange n objects into
• Combinations: nk = k!(n−k)!
k bins. This notation also refers to the binomial coefficients and is read “n choose k”.
• Cardinality Operator: n(S) , counts the (possibly infinite) number elements in set S.
• We will use “word” to refer to all possible distinct orderings of a group of letters
without concern for whether or not these orderings would show up in a dictionary.
Problems
1. There are 16 students in the Math club. How many different ways could they select a
president, vice president and treasurer for the club?
P (16, 3) =
16!
(16−3)!
=
16!
13!
= 16 × 15 × 14 = 3360
2. A website requires the user to choose a password with 5 letters and 2 numbers in that
order. Each letter or digit may be used only once. How many different passwords are
possible? How does this change if 5 letters and 2 numbers may be used in any order?
Let W be the set of all possible passwords.
n(W ) = P (26, 5) × P (10, 2) =
26! 10!
= 710, 424, 000
21! 8!
Part 2 is just a 36-letter alphabet from which we choose 7 items (without replacement):
n(W ) = P (36, 7) =
36!
≈ 4.20723 × 10−10
29!
3. The name of 11 students are placed in a hat. A teacher will reach into the hat and
select 3 names at one time. Each of those 3 students will win the same prize. How
many different groups of 3 winners could be chosen?
11!
Let G be the set of possible groups. Then n(G) = 11
= 3!8!
= 165
3
1
4. In how many ways can a sorority of 20 members select a president, vice president and
treasury, assuming that the same person cannot hold more than one office?
P (20, 3) = 20!
= 6840, assuming the officers are selected sequentially, that is, that the
17!
loser(s) from the Presidential race can enter the VP race, and so on.
5. Out of 7 consonants and 4 vowels, how many “words” of 3 consonants and 2 vowels
can be formed?
Let W be the set of “words” that are possible. If all 11 letters are distinct, then:
7 4
n(W ) =
7! = 1, 058, 400
3 2
Let’s identify the job each term performs:
• 73
from the group of 7 consonants, chooses 3 (unordered) to be included
• 42
from the group of 4 vowels, chooses 2 (unordered) to be included
• 7! Now that the letters to be used have been selected, we count the P (7, 7) ways
they can be permuted into different “words”
6. A license plate is to consist of three letters followed by two digits. How many different
license plates are possible if the first letter must be a vowel (a, e, i, o, u), and repetition
of letters is not permitted, but repetition of digits is permitted?
Let L be the set of all possible license plates we can make given the restrictions above.
We multiply by the symbols available at each step:
n(L) = (5)(25)(24)(10)(10) = 300, 000
7. A teacher is making a multiple-choice quiz. She wants to give each student the same
questions, but have each student’s questions appear in a different order. If there are
twenty-seven students in the class, what is the least number of questions the quiz must
contain?
Solution path: Guess-Check-Revise. The trick is to choose n such that P (n, n) ≥ 27.
If we guess P (4, 4) = 4! = 24, we realize the correct answer must be n = 5.
8. How many 10 digit numbers can be formed using 3 and 7 only?
Whatever we call the two symbols, {0, 1} or {3, 7} this is Base 2, and we can form
210 = 1024 binary numbers with 10 digits.
9. Everyone shakes hands with everyone else in a room. Total number of handshakes is
66. What is the total number of people in the room?
For n = 2, there is 1 handshake. For n = 3, the first person shakes with 2 folks, the
next person shakes hands with one other person. Continuing the pattern in a table:
n=2
n=3
n=4
n=5
..
.
n
1
1+2
1+2+3
1+2+3+4
..
.
=
=
=
=
1
3
6
10
..
.
1 + 2 + · · · + n = 66
So n = 12. I hate Guass’s formula, so I won’t include it. You may google it, if you’d
like. I prefer the Guass Idea to solve for n, which I will happily share with you some
time. If you don’t know either, you can continue the pattern in the table for a while
(or guess-check-revise).
10. In a college football training session, the defensive coordinator needs to have 10 players
standing in a row. Among these 10 players, there are 1 freshmen, 2 sophomores, 4
juniors, and 3 seniors, respectively. How many different ways can they be arranged in
a row if only their class level will be distinguished?
The question restated is, “How many distinct permutations of the ten letters FSSJJJJRRR
(10,10)
= 12, 600.
exist” (where we use “R” for “Senior”)?. Then n = P 2!4!3!
11. In one year, three awards (research, teaching, and service) will be given for a class of
25 graduate students in a statistics department. If each student can receive at most
one award, how many possible selections are there?
Order matters, so P (25, 3)
12. A president and a treasurer are to be chosen from a student club consisting of 50
people. How many different choices of officers are possible if:
(a) There are no restrictions.
P (50, 2) = 50 × 49
(b) A will serve only if he is president.
P (49, 2) + 49 = 492
(c) B and C will serve together or not at all.
P (48, 2) + 1
13. A developer of a new subdivision offers a prospective home buyer a choice of 4 designs,
3 different heating systems, a garage or carport, and a patio or screened porch. How
many different plans are available to this buyer?
n = 3 × 2 × 2 = 12
14. In how many different ways can a true-false test consisting of 8 questions be answered?
Whether we call the symbols {0, 1} or {T, F }, this is Base 2. There are 28 = 256
binary numbers of length 8.
15. Find the number of ways to arrange 16 items in groups of 4 at a time (order matters).
Let G be the number of different groupings possible. We use cascading permutations
to produce the correct answer, which is in a (somewhat) surprising form when we
simplify:
n(G) =P (16, 4) × P (12, 4) × P (8, 4) × P (4, 4)
16! 12! 8! 4!
=
12! 8! 4! 0!
=16!
=P (16, 16)
Think about why this is true. Does it make sense?
16. Find the number of ways to take 20 objects and arrange them in groups of 5 at a time
where order does not matter?
20 15 10 5
n(G) =
5
5
5
5
20!
15!
10!
5!
=
15!5!
10!5!
5!
5!0!
20!
=
5!5!5!5!
This is a multinomial coefficient (see last 5 problems on last page of this document).
17. How many ways are there to select a subcommittee of 7 members from among a committee
of 17?
17
7
18. How many ways are there to have a license plate with 3 letters followed by 4 numbers
in which the 4 numbers are listed from least to greatest (examples: 1234 or 0569)?
If we can calculate the quantity of 4-digit numbers that have a strictly increasing
pattern, we would multiply the result by 263 to the total number of license plates
possible.
Having never attempted something like this before, my first guess is to try selecting
four different digits, and note that only one of the 4! arrangements of these four digits
is ordered from least to greatest. Thus:
10
n(L) = 26
4
3
19. How many full house combinations are there in 5-card poker?
13 2 4 4
n(FH) =
2
1 3 2
Let’s point out the function of each term:
• 13
choosing 2 of the 13 card values to have in our hand, say J’s and 8’s
2
• 21
choosing the value (from the 2 in my hand) that will be the trips
• 43
choosing the trips
• 42
choosing the pair
20. From a group of 7 men and 6 women, five persons are to be selected to form a committee
so that at least 3 men sit on the committee. In how many ways can it be done?
Let M be the variable indicating the number of men selected in the 5-person group, so
that the event set M = 3 is a committee with exactly 3 men. Then we have:
7 6
n(M = 3) =
= 525
3 2
7 6
n(M = 4) =
= 210
4 1
7 6
n(M = 5) =
= 21
5 0
=⇒
n(M ≥ 3) =525 + 210 + 21 = 756
21. In how many different ways can the letters of the word ’LEADING’ be arranged in
such a way that the vowels are together?
We first use P (5, 5) = 5! to count the orderings possible, treating the vowels as a single
unit. We multiply this by P (3, 3) = 3! to account for the different orderings within the
group of 3 vowels. Thus, of the 5040 orderings, 720 of them have the vowels together.
22. Ten people go to a party. How many different ways can they be seated at a round
table?
P (9, 9) = 9! = 362, 880
23. There are 10 marbles in a bag numbered from 1 to 10. Three marbles are selected at
random without replacement. How many different ways are there of selecting the three
marbles?
P (10, 3) = 720
24. How many ways can I give 3 tin cans to 8 people if each person receives at most one
can?
P (8, 3)
25. An identification code is to consist of seven letters followed by three digits. How many
different codes are possible if repetition is permitted?
n(C) = 267 × 103
26. An 8-bit binary word is a sequence of 8 digits, of which each may be either a 0 or a 1.
How many different 8-bit words are there?
28 = 256
27. Jane and Thomas are among the 8 people from which a committee of 4 people is to
be selected. How many different possible committees of 4 people can be selected from
these 8 people if at least one of either Jane or Thomas is to be selected?
The simplest approach mathematically is calculate the committees that
contain neither
8
Jane or Thomas, and subtract from the total possible. We have 4 total committees
and 64 committees that exclude Jane and Thomas. Thus the number of committees
that include one or both of them is equal to 70 − 15 = 55.
28. Georgia license plates consist of three letters followed four numbers. How many license
plate combinations are possible if the license plate contains no vowels (A, E, I, O, U,
or Y) nor contains the number 5.
29. A company issues a questionnaire whereby each employee must rank the 5 items with
which he or she is most satisfied. The items are wages, work environment, vacation
time, job security, supervisors, health insurance, break time, and retirement plan. The
ranking is to be indicated by the numbers 1, 2, 3, 4 and 5, where 1 indicates the item
involving the greatest satisfaction and 5 the least. In how many ways can an employee
answer this questionnaire?
If I understand the problem, the employee is selecting 5 items to rate from 8 choices.
Then those chosen are ordered from 1 to 5. This is reminiscent of the “how many 4digit numbers have the digits in strictly ascending order” question. We will pick the 5
items with 85 , then produce all the possible orderings of the 5 items with P (5, 5) = 5!.
Thus, we have 56 × 120 = 6720 possible ways.
30. A keypad lock has 10 different digits, and a sequence of 5 different digits must be
selected for the lock to open. How many key pad combinations are possible?
P (10, 5) = 10 × 9 × 8 × 7 × 6× = 30, 240
31. There are five women and six men in a group. From this group a committee of 4 is
to be chosen. How many different ways can a committee be formed that contain three
women and one man?
32. How many 1 pair combinations are there in a standard deck of 52 cards?
3
13 12 4 4
n(1 Pair) =
1
3
2 1
Let’s point out the function of each term:
• 13
choosing 1 of the 13 card values to have as our pair, say J’s
1
• 12
choosing 3 other values that will be in my hand, but will not match my
3
pair value or each other
• 42
choosing the pair
3
• 41
choosing the single cards
33. A class of 30 students enter a room with 40 desks. How many different ways can those
students be seated?
34. If a school has lockers with 50 numbers on each combination lock, how many possible
combinations using three numbers are there?
35. From a group of 7 men and 6 women, five persons are to be selected to form a committee
so that at least 3 men are on the committee. In how many ways can it be done?
36. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can
be formed?
37. There are 6 boxes numbered 1, 2, ... 6. Each box is to be filled up either with a red
or a green marble in such a way that at least 1 box contains a green marble and the
boxes containing green marbles are consecutively numbered. The total number of ways
in which this can be done is?
38. How many integers, greater than 999 but not greater than 4000, can be formed with
the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?
39. If a license plate has exactly 3 letters and 4 digits, how many license plates are possible
if there must be exactly one vowel (excluding Y) and one 7 in the sequence?
40. Proteins are made up of chains of amino acids. Insulin is a relatively small protein
with 53 amino acid residues. How many possible proteins of length 53 can be made
with 20 possible amino acids for each position in the protein?
41. We wish to know how many color combinations can be made from four different colored
marbles if we use only three marbles at a time. The marbles are colored red, green,
white, and yellow.
42. Two girls and their dates go to the drive-in, and each wants a different flavored ice
cream cone. The drive-in has 24 flavors of ice cream. How many combinations of
flavors may be chosen among the four of them if each one selects one flavor?
43. We want to paint three rooms in a house, each a different color, and we may choose
from seven different colors of paint. How many color combinations are possible for the
three rooms?
44. Find the number of ways to take 10 people and place them in 3 groups of 3 where
order does not matter.
45. Find the number of ways to take 20 objects and arrange them in groups of 5 at a time
where order does not matter.
46. Find the number of ways to arrange 5 objects that are chosen from a set of 7 different
objects given that order matters.
47. What is the total number of possible 5-letter arrangements of the letters WHITE if
each letter is used exactly once?
48. How many different 3-digit numerals can be made from the digits {4, 5, 6, 7, 8} if a digit
can appear just once in a numeral?
49. We wish to know how many color combinations can be made from four different colored
marbles if we use only three marbles at a time. The marbles are colored red, green,
white, and yellow.
This is equivalent to a 3-letter “word” created using an alphabet with only 4-letters.
There are 43 possibilities.
50. At the pizza place, there are 8 toppings that you can put on your pizza. If you can
order any number of those 8 toppings, then how many different toppings could you
possibly order?
5 Multinomial Counting Problems
Formulas and Notation
• Multinomial Coefficients: k1 ,k2n,...,km = k1 !k2n!!···km ! , given unordered groups of sizes
P
k1 , k2 , . . . , kn such that
ki = n, the multinomial coefficient gives the number ways
to arrange n objects into these groups.
• If this is confusing, a string of binomial coefficients can be used instead, choosing each
group one at a time rather than choosing all the groups at once.
Problems
1. Ten consultants will be divided into three teams of three and leader. How many
different possibilities for the team configurations exist?
Let G be the set of all possible groupings that meet these requirements. We can use
the multinomial coefficient, as follows, with the expressing mapped back to the same
process with a cascading string of binomial coefficients:
10
n(G) =
3, 3, 3, 1
10!
=
3!3!3!1!
10! 7! 4! 1!
=
7!3! 4!3! 3!1!
1! 10 7 4 1
=
3
3 3 1
2. A teacher divides a class of 20 students into five teams of 4. How many different
possibilities exist?
Let G be the set of all possible groupings that meet these requirements. We can use
the multinomial coefficient, as follows, with the expressing mapped back to the same
process with a cascading string of binomial coefficients:
20
n(G) =
4, 4, 4, 4, 4
20!
=
4!4!4!4!4!
20! 16! 12! 8! 4!
=
16!4!
12!4!
8!4! 4!
4! 20 16 12 8 4
=
4
4
4
4 4
3. A teacher divides a class of 20 students into four teams of 5. How many different
possibilities exist? Compare this to your answer to the previous question. Are there
more or fewer possibilities than before?
4. A dormitory hallway of 25 girls is selecting several groups to host socials. They need
two groups of 4 girls, two groups of 3 girls, two groups of 2 girls and a group of 7 girls
for the semester finale. How many possibilities exist?
Let G be the set of all possible groupings that meet these requirements. We can use
the multinomial coefficient, as follows:
25
n(G) =
4, 4, 3, 3, 2, 2, 7
25!
=
4!4!3!3!2!2!7!
5. Compare the expressions. Calculate the total number of unique orderings of the letters
DREADED. Then calculate the number of ways seven boys can be divided up into a
team of 3, a team of 2 and a pair of boys to not be on either team.