Section 5.1: The Basics of Counting: Discrete Mathematics and Its Applications: Page - 2

Rosen, Discrete Mathematics and Its Applications, 6th edition

Extra Examples

Section 5.1—The Basics of Counting

p.340, icon  at Example 14

#2.   Find the number of strings of length 10 of letters  of the alphabet, with no repeated letters,  that begin with a vowel.

Solution:

Keep in mind a row of ten blanks _ _ _ _ _ _ _ _ _ _ _:  .

There are five ways in which the first letter in the string can be a vowel. Once the vowel is placed in the first blank,  there  are 25 ways in which to fill in the  second blank,  24 ways to fill in the third  blank,  etc.  Using the extended  product rule we obtain

5     •    25 • 24 • 23 ••• 18 • 17   .

place vowel  place  other letters

p.340, icon at Example 14

#3.   Find  the number  of strings of length  10 of letters  of the alphabet, with no repeated letters,  that have

C and V at the ends (in either  order).

Solution:

Using a row of ten blanks, we first count the number  of ways to have the pattern

  C _ _ _ _ _ _ _ _ V .

The number  of ways to fill in the eight interior letters  is 24 • 23 ••• 18 • 17. Similarly, the number  of words of the form

  V _ _ _ _ _ _ _ _  C

is 24 • 23 ••• 18 • 17.

Therefore,  by the sum rule the answer is (24 • 23 ••• 18 • 17)  + (24 • 23 ••• 18 • 17) = 2(24 • 23 ••• 18 • 17).

p.340, icon at Example 14

#4. Find the number of strings of length 10 of letters of the alphabet, with repeated letters allowed, that

have vowels in the first two positions.

Solution:

Keep in mind a row of ten blanks:                                                  .

If vowels must be in the first two positions and letters  can be repeated,  we obtain the product

5 · 5 · 26 · 26 ··· · 26 ,

 which is 52  · 268.

p.340, icon  at Example 14

#5.   Find  the number  of strings of length  10 of letters  of the alphabet, with no repeated letters,  that have vowels in the first two positions.

Solution:

Keep in mind a row of ten blanks:                                                   .

We will first count the number  of ways to place vowels in the first two blanks.  We can choose any of the five vowels for the first blank and any of the remaining four vowels to put  in the second blank.  Because we must

do both,  there are 5 · 4 = 20 ways to place the two vowels.

Next  we will place eight  of the remaining  24 letters  in the remaining  eight  blanks.   This  can be done in

24 · 23 ··· 18 · 17 ways.

Therefore,  by the  product  rule, the number  of ways to place vowels in the first two blanks and eight letters in the remaining  eight blanks is

(5 · 4) · (24 · 23 ··· 18 · 17) .