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212
Solutions to Review Exercises
2. As a group, the girls can be arranged in 6! ways. Then, together with the boys, they form 8 "units" to
be arranged in a row. Since the girls cannot appear first, there are seven choices for the first position in
the row. Then there are six choices for the last position (again, not the girls). There are 6! ways to fill
out the remaining positions, so the answer is 7(6)(6!)(6!)
=
21,772,800.
3. The answer is zero! In a circle, if the six girls are together, then the boys must also be together.
4. Let
A
be the set of permutations containing
aeb
and
B
the set of permutations containing
bef.
We
want
IA
UBI.
By the InclusionExclusion Principle, this is
IAI
+
IBIIA
n
BI.
Now
IAI
=
4!
(aeb
is
a single unit in a string of four "units"),
IBI
=
4! and
IA
n
BI
=
2! (if both
aeb
and
bef
appear, then
aebef
must appear). Thus
IA
U
BI
=
IAI
+
IBI
IA
n
BI
=
24
+
24  2
=
46.
5. (a) Let
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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