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Here are some more counting examples along with brief explanations on how to get the
correct answers. The first is taken from the quiz on December 1. Since there were several
people who had trouble with it, please bring your questions to class if this explanation doesn’t
clear things up.
Example 1: Consider the set A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21}. How many subsets of A
a. contain exactly seven numbers?
C(11, 7)
Since sets are unordered this just asks how many ways are there to choose 7 numbers from a
set with 11 numbers
b. contain all multiples of 3 (not only the multiples of 3)?
27
There are four multiples of 3 and a subset must contain them plus any subset of the remaining 7
numbers. Thus, the problem becomes how many subsets can you make from a set of size 7.
c. contain exactly two prime numbers? C(7, 2)•24
There are seven prime numbers in the set: 3, 5, 7, 11, 13, 17, and 19 so there are C(7, 2)
ways to choose the primes and any or all of the four composite numbers may also be in the
set so the answer is the product C(7, 2)•24
d. contain exactly six elements, exactly three of which are less than 10? C(5, 3)•C(6, 3)
Choose three of the five numbers less than 10 and three of the six numbers greater than 10.
Example 2: A committee of size 5 is chosen from a club with 20 males and 25 females. Find
the number of committees in which
a. a female is designated as the chair and the committee has 3 males and 2 females.
C(20, 3)•C(25, 2)•2 (Choose the three females, choose the two females, and then select
on of the two chosen women to be the chair.) Alternatively, choose the three males and
then choose the woman who will be chair and then choose the other woman. C(20,
3)•25•24
b. One is designated as the chair and at least one is a male.
First find the number of 5 member committees with no males and subtract that from the
number of possible committees to get the number of 5 member committees with at least
one male. Then, choose one of the five committee members to be chair giving:
[C(45, 5) - C(25, 5)]•5
c. One is designated as the chair and exactly one male is on the committee.
Choose the one male, choose four females, and then pick the chair from the five people
selected to get: C(20, 1)•C(25, 4)•5
Example 3: More variations of the powerball problem
a. How many powerball tickets are there in which no number is a multiple of 5?
When the numbers on the white balls are considered, there are 55 – 11 = 44 balls whose
numbers are not multiples of 5. Similarly, there are 42 – 8 = 34 numbers on the red balls
that are not multiples of 5. Since the white balls can be chosen in C(44, 5) ways the total
number of tickets satisfying this condition is C(44, 5)•34.
b. How many ways are there to win a lesser prize? Let’s consider a few of the possibilities—
for practice you can calculate the others.
1) How many tickets match only the powerball?
Since the powerball must be matched, there is only one way this can be done. However,
even though none of the white balls match, there must still be five white ball numbers
chosen. This may be done in C(50, 5) ways.
2) How many tickets will match exactly three white balls and not the powerball?
Of the five numbers chosen, you match exactly three which can be done in C(5, 3) ways.
Since you must choose a total of five numbers and the other two do not match, there are
C(50, 2) ways to choose the nonwinning numbers. There are 41 ways in which the
powerball is not matched, so the number if tickets is C(5, 3)•C(50, 2)•41.
3) On how many tickets are all numbers chosen even?
There are 27 even numbers on the white balls and 21 even-numbered red balls, so the
answer is C(27, 5)•C(21, 1).
Example 4: How many distinct arrangements are there of the letters in the word
SUCCESSFUL?
10!/3!•2!•2!
There are 10! was to arrange the 10 letters, but because there are three S’s, two C’s and two
U’s we need to divide by the factorials of each of those numbers to eliminate double counting.
Example 5: A candy shop has 10 different kinds of chocolates. In how manh ways can four
chocolates be chosen if
a. all must be different kinds C(10, 4)
b. there are no restrictions C(13, 4)
In this case, we may have from 0 to 4 pieces of chocolate for any particular kind. Thus, we
are allowing repetition. The number of was this can be done is C(4 + 10 – 1, 4) = C(13, 4).
Note: In class I was using the n and r slightly differently from they way they are used in the
text. The authors use the formula C(r + n-1, n) and I was using C(n + r-1, n) which is clearly
the same value.
The main thing to remember is that the “bottom” number in the binomial
coefficient is the number of things being selected.
I didn’t explicitly cover this in class, but here’s another way to find solutions to some counting
problems when exhaustive examination of all possibilities seems to be the approach to use.
Decision trees: Sometimes the only way to count something is to draw a decision tree in
which each leaf represents a decision. (i.e. exhaustively find all of the possible ways to
succeed.) For example, consider a best 3 of 5 tournament in which the teams A and B are
evenly matched. We want to know in how many ways the tournament can be played. We’ll
use the convention that if A wins we move to the left and if B wins we move to the right. It can
be seen that in this case there are 20 ways for the tournament to be played.