ICS 6D ( Discrete Mathematic): Solution to Homework #6
Section 5.1
4.
12
*
2
*
3
8.
Here’s one way to count:
Take the number of all 3letter initials, 26
3
, and substruct the number of
3letter initials that have one repetition, which is 3
*
26
*
25 (because there’s three possibilities for where
the nondoubled letter is, 26 choices for the doubled letter, and 25 choices for the nondoubled letter), and
3letter initials, which is 26. So the answer is 26
3

(3
*
26
*
25 + 26).
20.
(a)
999
/
7 .
1
(b)
999
/
7

999
/
77 , because you have to substruct numbers that are divisible by
both
7 and 11, which
is the same set as the set of numbers that are divisble by 77.
(c)
999
/
77 , see above.
(d)
999
/
7
+
999
/
11

999
/
77 , using the inclusionexclusion principle and using part (c).
(e) ( 999
/
7

999
/
77 ) + ( 999
/
11

999
/
77 ) =
999
/
7
+
999
/
11

2 999
/
77 .
It’s the same
argument as in part (b), but used twice, first for all numbers divisible by 7 and not 11, and then for all those
that are divisible by 11 but not by 7.
(f) 999

( 999
/
7
+
999
/
11

999
/
77 ), by part (d).
(g) 9 + 9
*
9 + 9
*
9
*
8, summing the counts of numbers with no repetitions among digits which have,
respectively, exactly one digit, exactly two digits, and exactly three digits. Note that in each case the first
digit has to be nonzero, so there is 9 choices, and the other digits can be any digit among all 10 digits except
of the previously taken digits, so there are 9 choices of the second digit, 8 choices of the third digit, etc.
(h) This one was tricky! As I went over it in the class, there’s many incorrect paths one can fall into,
but here’s a way that works: Let
o
i
be the count of
i
digit numbers that satisfy this constraint, i.e. they
are even and all their digits are different. The answer is
o
1
+
o
2
+
o
3
. We have that
o
1
=
{
2
,
4
,
6
,
8
}
= 4.
We also have
o
2
= 4
*
8 + 1
*
9, because if the number has two digits then if its last digit is nonzero (4
choices) then its first digit has 8 choices (any digit except zero and the one that’s equal to the last digit),
and if its last digit is zero (1 choice) then its first digit has 9 choices (any digit except zero). Finally we have
o
3
= 4
*
8
*
8 + 1
*
9
*
8 because if the number has three digits then if its last digit is nonzero (4 choices)
then its first digit has 8 choices (any digit except zero and the one that’s chosen as the last digit) and its
middle digit has also 8 choices (any digit except of the two that are chosen as first and last digit), and if its