6.2#11
a)
A bit string is a sequence of 0’s and 1’s. How many bit strings have
length 8?
28 = 256.
b)
How many bit strings of length 8 begin with a 1?
Half of these: 27 = 128.
c)
How many bit strings of length 8 both begin and end with a 1?
Half of the bit strings end with a 1: 27 = 128 and half of these
begin with a 1 so: 26 = 64..
6.2#12
a)
How many hexadecimal numbers are there from 1016 through FF16?
Recall that an inclusive count is m – n + 1.
(15*161 + 15*160) - (1*161 + 0*160)
+ 1 = (240 + 15) – (16) + 1 = 240
Or using the multiplication rule we have 1 through F choices for the first digit
and 0 through F choices for the second digit, each choice independent of the
other, so there are 15*16 = 240 choices in all.
b)
How many hexadecimal numbers are there from 5016 through FF16?
(15*161 + 15*160) - (5*161 + 0*160)
+ 1 = (240 + 15) – (80) + 1 = 176
Or using the multiplication rule we have 5 through F choices for the first digit
and 0 through F choices for the second digit, each choice independent of the
other, so there are 11*16 = 176 choices in all.
c)
How many hexadecimal numbers are there from 3016 through AF16?
(10*161 + 15*160) - (3*161 + 0*160)
+ 1 = (160 + 15) – (48) + 1 = 128
Or using the multiplication rule we have 3 through A choices for the first digit
and 0 through F choices for the second digit, each choice independent of the
other, so there are 8*16 = 128 choices in all.
6.2#20
a)
How many integers are there from 100 through 999?
999 – 100 + 1 = 900
Or using the multiplication rule we have 9 ways to count the first digit, 10
ways to count the second digit and 10 ways to count the third digit. 9*10*10 =
900.
b)
How many odd integers are there from 100 through 999?
101 + 103 + … + (2k + 1), k = 50 … 499. There are 499 – 50 +
1 = 450 odd integers from 100 through 999.
Or using the multiplication rule we have 5 ways to count the third digit (this
is what makes the numbers odd), 9 ways to count the first digit and (can't use
0), 10
ways to count the second digit. 9*10*5 =
450.
c) How many integers from 100 through 999 have distinct digits?
Using the multiplication rule we have 9 ways to count the first digit (can't use
0), 9 ways
to count the second digit (can't use the digit used in the first one) and 8 ways to count the third digit (can’t use
the second or third digits). 9*9*8 = 648.
d) How many odd integers from 100 through 999 have distinct digits?
Using the multiplication rule we have 5 ways to count the third digit (this is
what makes the numbers odd), 8 ways to
count the first digit (can't use 0 or the third digit) and 8 ways to count the
second digit (can’t use the first or third digits). 8*8*5 = 320.
e)
What is the probability (A) that a randomly chosen three-digit integer
has distinct digits? (B)has distinct digits and is odd?
P(A) = 648/900 ®
.72 = 72%
P(B) = 320/900 ®
.356 = 35.6%