184
Solutions to Exercises
11. Let
A
be the set of permutations which contain the pattern
bge:
IAI
=
5! (thinking of
bge
as a single
symbol). Similarly, if
B
is the set of permutations which contain
eaj,
IBI
=
5!. For a permutation
to contain both
and
eaj
it must contain
bgeaj,
so
IA n
BI
=
3!. The number containing either
or
is
IA
U
=
+ IBI
 IA
n
=
5!
+
5! 
3!
=
234. Hence, the answer is
7! 
234
=
5040 
234
=
4806.
12. [BB] Let
N
be the set of lines in which the Noseworthys are beside each other and
A
the set of lines
in which the Abbotts are beside each other.
(a) For each of the two ways in which the Noseworthys can stand beside each other, there are 3!
possible linesthe Noseworthys as one unit and the two Abbotts. Thus,
INI
=
2 x 3!
=
12.
(b) The number of possible lines altogether is 4!
=
24. Thus,
INcl
=
24 
12
=
12.
(c) There are 2! orders for the couples and, in each case, two arrangements for the Noseworthys and
two for the Abbotts:
IN
n AI
= 2 x 2 x 2 =
8.
(d) The Noseworthys must be in positions 2 and 3 in either order; the Abbotts must be in positions 1
and 4 in either order. There are four lines.
This is the end of the preview. Sign up
to
access the rest of the document.
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
 Permutations, Graph Theory

Click to edit the document details