MAT 344
Spring, 1999
94
Exercise: The number of derangements D
n
Let a
i
be the property that i is in its natural position in the permutation
B
,
i.e.
B
(i) = i.
Then D
n
is the no. of perms with none of the properties a
i
.
Then N = n!, N(a
i
) = (n - 1)!, N(a
i
a
j
) = n - 2)!
= (n- r)! The number of r-tuples is
.
Thus,
N
a
i
1
a
i
2
,
,a
i
r
k
r
D
n
= N(a
N
1
a
N
2
a
N
n
) =
j
n
r
’
0
(
1)
r
n
r
(n
r)!
Exercise: (Euler Phi Function
) Let n =
p
e
1
1
p
e
2
2
±
p
e
k
k
Denote by
N
(n) the number of integers
from 1 to n (incl.) rel. prime to n.
Find
N
(n).
Let a
i
be the property that an integer in [n] is divisible by p
i
.
Then
N
(n) =N(a
N
1
a
N
2
a
N
k
).
Here
N = n, N(a
i
) =
, etc.
n
p
i
, N(a
i
a
j
)
’
n
p
i
p
j
Thus,
N
(n) = n -
j
i
n
p
i
%
j
i
1
²
j
2
n
p
i
1
p
i
2
j
n
p
i
1
p
i
2
p
i
3
= n
1
1
p
1
1
1
p
2
1
1
p
k
Generalization
Suppose we want the number of objects which have exactl
y
m of the properties (any m of the r
properties will do, m
0
[0,r].
Let e
m
be this number. Let s
m
be the number of objects
s
m
=
j
N
a
i
1
a
i
2
,
,a
i
m
where the sum is taken over all choices of m distinct properties
.
a
i
1
,
,a
i
m
N.B.
s
m
counts elements more than once.
Every element counted by s
m
has at least
m properties but
those with more than m properties get counted many times.
For example, if an object has the
properties a
1,
a
2
,
, a
m + 1
, it gets counted in N(a
1
a
2
a
m
),
N(â
1
a
2
a
3
a
m + 1
), N(a
1
â
2
a
3
a
m + 1
) etc.
8