Section 7.6
7'
(1

.1.
+
.1.

.1.
+
.1.
_
.1.
+
.1.
_
.1.)
.
I!
2!
3!
4!
5!
6!
7!
7!

7!
+ 7 . 6 . 5 . 4 . 3 
7 . 6 . 5 . 4 + 7 . 6 . 5 
7 . 6 + 7 
1
2520 
840 + 210 
42
+ 6
=
1854
Ds
8'
(1

.1.
+
.1.
_
.1.
+
.1.
_
.1.
+
.1.
_
.1.
+
.1.)
.
I!
2!
3!
4!
5!
6!
7!
S!
8!

8!
+ 8 . 7 . 6 . 5 . 4 . 3 
8 . 7 . 6 . 5 . 4 + 8 . 7 . 6 . 5 
8 . 7 . 6
+8·
7 
8 + 1
=
20,160 
6720 + 1680 
336 + 56 
7
=
14,833
2.
This is
D
26
=
26!(1
fr
+
tr

lr
+
...
+
2~!)'
3.
[BB] This is
Dn
=
11!(1 
fr
+
ft

lr
+ ... 
1~!)'
4.
D50
5.
(a)
D7
6.
(a)
D
20
(b)
[BB]
7!

D7
(b)
20!

D
20
(c)
1
201
(c) [BB] There are
20
choices for the person who receives his or her own hat. For each choice, the
19
other people can get their hats in
D
19
ways. Hence, the answer is
20D
19
.
(d) This is the answer to
(b)
less the answer to (c); that is,
20!

D
20

20D
19
.
(e) This means either no person receives hislher own hat or exactly one person receives hislher own
hat or exactly two people receive their own hats. The answer is
D
20
+
20D
19
+
e2
0
)D
1S
.
7. (a) [BB] Let
Al
be the set
of
permutations
of
19
such that
1
is in position
1;
A3
the set
of
perm uta
tions
of
19
such that 3 is in position
3.
Define
A5, A7
and
A9
similarly. Then
I Ui
Ail
=
Li
IAil
Li<i
IAi
nAil
+
Li<i<k
IAi
n
Ai
n
Akl
 Li<i<k<t
IAi
n
Ai
n
Ak
n
Atl
+
IAI
n
A3
n
A5
n
A7
n
A91
=
5(8!) 
(~)7!
+
(~)6!

(~)5!
+
4!.
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 Summer '10
 any
 Permutations, Graph Theory, Natural number, Prime number, Quantification, Bijection

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