ICS 6D ( Discrete Mathematic): Solution to Homework #7
Section 5.3
Disclaimer:
There are usually many ways to count the same thing, so here are just some ways and you
might come up with a diﬀerent one, just as good. Although be critical and if you come with something else,
make sure it comes out to the same thing, and try to understand why or why not!
2.
P
(7) = 7!
6.
You have to compute these yourself!
8.
P
(5), it’s the number of orderings, i.e. permutations, among 5 elements.
12.
a)
C
(12
,
3), number of 3element subsets in 12element set, because there are 12 indices and you have to
pick exactly three where you’ll place a one.
b)
C
(12
,
0) +
C
(12
,
1) +
C
(12
,
2) +
C
(12
,
3), treating bitstrings with exactly zero, one, two, or three ones
independently.
c)
∑
12
i
=3
C
(12
, i
), which is equal to 2
12

(
C
(12
,
0) +
C
(12
,
1) +
C
(12
,
2)).
d)
C
(12
,
6), because there has to be exactly six ones (and six zeros).
16.
You can just add them up:
C
(10
,
1) +
C
(10
,
3) +
C
(10
,
5) +
C
(10
,
7) +
C
(10
,
9).
You might also be able to do something more clever by comparing the oddelement subsets and even
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 Fall '07
 Jarecki
 Logic, Bijection, Discrete Mathematic

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