Discussion 11 and 12
Spring 2015
weeks of April 6 and April 13
This notes on counting will last at least two weeks. I’ve supplied answers for several examples,
but I’d suggest you try to work out the problem first and see if your answer agrees.
Example 1: One hundred tickets numbered 1, 2, 3, ..., 100, are sold to 100 different people for a
drawing. Four different prizes are awarded, including a grand prize. How many ways are there
to award the prizes if
1). there are no restrictions? Answer: 100·99·98·97 ways.
2) the person holding ticket 47 wins the grand prize? Answer: 1·99·98·97 ways
3) the person holding ticket 47 wins a prize? Answer: 4·99·98·97 ways
4) the person holding ticket 47 doesn't win a prize? Answer: 99·98·97·96 ways
5) people holding tickets 19 and 47 both win?
Answer: 4·3·98·97 ways since 19 may win any of the four prizes, 47 wins one of the
remaining three prizes, and two of the other 98 ticket holders win the other two prizes.
6) people holding tickets 19, 47, and 73 all win? Answer: 4·3·2·97 ways
7) people holding tickets 19, 47, 73 and 97 all win? Answer: 4·3·2·1 ways
8) none of the people holding tickets 19, 47, 73 and 97 wins? Answer: 96·95·94·93 ways
9) the grand prize winner is holding ticket 19, 47, 73, or 97? Answer: 4·99·98·97 ways
10) people holding 19 and 47 win prizes and those holding 73 and 97 do not?
Answer: 4·3·96·95 ways
Example 2: There are eight books in the library on a particular topic which you are studying.
a. How many ways are there to arrange the books on a shelf?
b. How many ways are there of borrowing one or more of these books?
Example 3: How many four digit numbers contain at least one even digit? Remember that a
four-digit number cannot have 0 as the first digit.
Ans. total number - # with all odd digits = 9000 – 54
Example 4: How many four digit numbers with all digits distinct contain exactly two even digits,
no repetition allowed?
Ans. Let's split this up into two cases: even in first position or odd in first position
even in first position: 4 choices for the even digit in first position (can’t have 0)
3 places for second even digit
4 choices for second even digit
5·4 ways to put in the odd digits
Total: 3·43·5
odd in first position:
5 choices for the odd digit in first position
3 places for second odd digit
4 choices for second odd digit
5·4 ways to put in the even digits
Total: 3·42·52
The answer is the sum of these two values.
(Note: this problem is much more difficult if repetition is allowed since you have to avoid double
counting.)
Example 5: A test contains 12 questions, with six questions each in parts A and B. A student
must answer 7 questions, but cannot answer more than 5 from one part. How many ways can
the questions to answer be chosen? (Keep in mind we’re interested in the set of questions
chosen from each part not the specific questions.)
Example 6: A class has 20 male and 25 female students. How many ways can a committee of
5 students be selected so that a male is designated as the chair and the remaining members are
2 males and 2 females?
Ans. 20·C(19, 2)·C(25,2)
This answer reflects choosing the male to be the chair and then the remaining members. This is
the same as 3·C(20,3)·C(25,2) which corresponds to choosing three men and two women and
then choosing one of the three men as the chair.
Example 7: A committee is chosen from a group containing 4 men and 6 women.
a)
b)
c)
d)
A committee of size three can be found in (10·9·8)/ 3! = C(10, 3) ways.
A committee of size three with 1 man and 2 women can be found in C(4,1)·C(6,2) = 60 ways.
A committee with two men and three women can be found in C(4, 2)·C(6, 3) = 120 ways.
A committee of size 4 with at least as many women as men can be found in
C(4,0)·C(6,4) + C(4, 1)·C(6,3)
+ C(4,2)·C(6,2)
ways.
all women
+ 1 man and 3 women + 2 men and 2 women.
Example 8: Suppose you have a 10 question multiple choice test with four possible answers for
each question: a, b, c, and d. Assume all questions are answered. Let's count the number of
answer keys under several different conditions:
a) If there are no restrictions, the number of answer keys is 410.
b) If there are exactly three a's and exactly two b's, here's the way to get the answer:
Choose the positions for the a's. There are C(10, 3) was to do this.
Choose the positions for the b's. There are C(7, 2) ways to do this.
Fill the remaining places with c's and d's. This can be done in 25 ways.
Thus, the answer is the product of these: C(10, 3)·C(7, 2)·25.
c) At least two answers are a's. There are two ways to do this:
exactly two a's + exactly 3 a's +  + exactly 10 a's. The better way is to use subtraction:
Total number of answer keys - those with no a's - those with exactly one a
410
- 310
- 10·39
The following approach will not work for part c): Choose the positions for two a's. Then, fill the
remaining 8 spots with any of the 4 other possibilities. The problem with this is that it causes
double counting. For example, suppose you choose to answer questions 1 and 2 with a and
then arbitrarily put in the rest of the answers to get:
a a b c d a b c d d.
Suppose instead you put a's in the first and 6th positions and get the following:
a a b c d a b c d d
The two keys are identical and would have been counted twice by this method.
Example 9: How many seven-card poker hands contain
a. three cards of one rank and two each of two other ranks 13·C(4, 3)·C(12, 2)·[C(4, 2)] 2
b. three cards each of two ranks C(13, 2)·[C(4, 3)]2·44
c. two cards each of three ranks C(13, 3)·[C(4, 2)] 3·40
d. four cards of one suit and three of another
e. three cards each of two suits and one card of a third suit?
f. exactly two queens, exactly four spades and no hearts?
Example 10: A hostess chooses 6 dinner guests from a list of 14 friends. In how many ways
can the guests be chosen if
a. six are boring and six are interesting and she wants to have at least one of each?
C(14, 6) – C(8, 6) – C(8, 6)
total – no interesting – no boring
b. two of her friends dislike each other and neither will come if the other is invited?
Interpretation: one or the other may be invited, but not both.
C(13, 6) + C(13, 6) – C(12, 6) exclude 1st person + exclude 2nd – exclude both
Alternatively this could be counted as
C(13, 5) + C(13, 5) + C(12, 6) include 1st person or include 2nd person + include neither
Example 11: Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Find the number of subsets of S that
a. contain the number 5 29
b. contain only even numbers 25 - 1 (Don't count empty set)
c. contain exactly three elements, one of which is 3
d. contain exactly five elements, but neither 3 nor 4
e. contain exactly one prime number
f. contain four numbers, exactly one of which is prime.
Example 12: A group consisting of 11 men and 7 women are to be seated together. In how
many ways can this be done if
a. they are seated in a row and all men sit together and all women sit together? 2•11!•7!
b. they are seated in a row and all men sit together? 8•11!•7!
(There are 8 possible seats for the first man to occupy)