Spring’2001
Assignment #1
Solution Key
#1 (p. 13, #18). There are seven different basic systems, four modems, three CD ROM
drives and six different printers. How many different 4 configurations can be made from
these elements?
Ans: 7
⋅
4
⋅
3
⋅
6=504
#2. (p. 13, #22) How many positive integers
n
can we form using the digits 3, 4, 4, 5, 5,
6, 7 if we want
n
to exceed 5,000,000?
Under the condition
n
>5,000,000 we have three choices for a digit in the leftmost
position. It may be 5, 6 or 7, which can be followed by any sequence of remaining six
digits.
Note, that in case 5,***,***
we need to count permutations of six digits with two
identical, 3, 4, 4, 5, 6, 7. This number is:
360
3
4
5
6
!
2
!
6
5
=
⋅
⋅
⋅
=
=
N
In case the rightmost digit is 6 we need to count permutations of six digits, among
which two are identical and another two are identical (3, 4, 4, 5, 7):
180
!
2
!
2
!
6
6
=
⋅
=
N
If
n
starts with 7, the number of
such integers will be the same,
180
7
=
N
.
By the rule of sum the total number of integers is:
720
7
6
5
=
+
+
=
N
N
N
N
#3. (p. 14, #34) How many distinct 4-digit integers can one make from the digits 1, 3, 3,
7, 7, and 8?
We need to count the 4-arrangements of 6 digits when two of them are of the
same kind and two other are of another kind.
Consider any arrangement been made in two steps: at the first step select four
digits from six and at the second step arrange them in some particular order. The number
of possible arrangements depends on how many identical digits are available in the
selection. So we can consider the following cases:
1)
All digits are distinct. There is only one selection of this kind {1, 3, 7, 8}.
There exist 4!=24 permutations of these digits.
2)
There are exactly two identical digits in a selection. It might be either 3 or 7
(but not both). For each of these subcases we need to count 2-selections from
remaining three digits. This number is
3
!
1
!
2
!
3
)
2
,
3
(
=
=
C
So, we have 2
⋅
3=6 selections with two identical digits. By the product rule this
number should be multiplied by the number of permutations (of four when two of them
are identical). This number is
12
!
2
!
4
=
.
1