Here are some more counting examples along with brief explanations on how to get the
correct answers.
Example 1: The English alphabet contains 21 consonants and 5 vowels. Consider strings of
length 6 all of whose letters are different.
a. How many such strings are there? Answer P(26, 6) = 26•25•24•23•22•21
b. How many of these contain exactly one vowel?
There are 5 ways to choose the vowel, 6 places to put the vowel, P(21, 5) ways to place
the consonants so the answer is the product of these numbers: 5•6•P(21, 5)
c. How many contain exactly two vowels?
This can be broken down into two parts—placing the vowels and placing the consonants.
There are C(6,2) ways to choose the positions of the vowels, 5•4 ways to put two vowels
into those positions, and finally P(21,4) ways to place the consonants. The final answer is
the product of these three numbers.
d. How many contain at least one vowel?
Take the number of strings of length 6 and subtract the number of strings consisting only
of consonants i.e. P(26, 6) – P(21, 6)
Example 2: The English alphabet contains 21 consonants and 5 vowels. Consider strings of
length 6 where letters may be repeated.
a. How many such strings are there? Since there are 26 choices for each position in the
string the answer is 266.
b. How many of these contain exactly one vowel?
There are 5 ways to choose the vowel, 6 places to put the vowel, and 21 5 ways to place
the consonants for a total of 5•6•215.
c. How many contain at least one vowel?
Take the number of strings of length 6 and subtract the number of strings consisting only
of consonants i.e. 266 - 216
Example 3: 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 4: 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 5: 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 59 – 11 = 48 balls whose
numbers are not multiples of 5. Similarly, there are 35 – 7 = 28 numbers on the red balls
that are not multiples of 5. Since the white balls can be chosen in C(48, 5) ways the total
number of tickets satisfying this condition is C(48, 5)•28.
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 (incorrect) white ball
numbers chosen. This may be done in C(54, 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(54, 2) ways to choose the nonwinning numbers. There are 34 ways in which the
powerball is not matched, so the number if tickets is C(5, 3)•C(54, 2)•34.
3) How many tickets will match exactly three white balls and the powerball.
Using the same kind of reasoning as in the first two part, there is only one way to match
the powerball and C(5, 3)•C(54, 2) ways to match exactly three white ball numbers so the
answer is C(5, 3)•C(54, 2)• 1.
4) On how many tickets are all numbers chosen even?
There are 29 even numbers on the white balls and 17 even-numbered red balls, so the
answer is C(29, 5)•C(17, 1).
Example 6: 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 7: A candy shop has 10 different kinds of chocolates. In how many 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 ways this can be done is C(4 + 10 – 1, 4) =
C(13, 4).
Note: The authors use the formula C(r + n-1, n) and I usually use 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.