哈工大深圳组合数学第六章习题答案

Solution: Let A1 be the number that a appear consecutively Let A2 be the number that b appear consecutively Let A3 be the number that c appear consecutively Let A4 be the number that d appear consecutively
1 1 1 1 Dn n!(1 ... (1) n ) , this will be the case that n = 4, 1! 2! 3! n! 1 1 1 1 4 So the result D4 C8 * 4!(1 ) = 630 1! 2! 3! 4!
y1 y2 y3 y4 13 0 y1 5 , 0 y4 4 . Let
0 y2 7 ,
0 y3 4 ,
S
be combination of positive solution for this
equation, let A1 be the combination that y1 >= 6, A2 be the combination that y2 >= 8, A3 be the combination that y3 >= 5, A4 be the combination that y4 >= 5.
4. Count the permutations i1i2i3i4i5i6 of {1, 2, 3,
4,
5, 6},
where i1 1,2,3 ; i2 1 ; i3 1 ; i5 5,6 ; and i6 5,6 .
Solution: Let Xi be what can be occur in position i, then X1={1,2,3}, X2={1}, X3={1}, X5={5,6}, X6={5,6}. Then we calculate the chessboard polynomial, the chessboard is like this:
7 A1 = C7 41 = 120 8 A3 = C8 41 = 165 5 A2 = C5 41 = 56 8 A4 = C8 41 = 165
A1 A2 = 0 let
u1 y1 6 ,
u2 y2 , u3 y3 5 , u4 y4
so u1 u2 u3 u4 2
R(
) = xR ( ) + R(
) = x 2 3x 1
R(
) = 2x2 4x 1
(*)
R(S) = 8x 4 26 x3 26 x 2 9 x 1 R6(C) = 6!-9*5!+26*4!-26*3!+8*2!-0*1! = 124
5. How many circular permutations are there of the multiset {2· a, 3· b, 4· c,5· d},where, for each type of letter, all letters of that type do not appear consecutively?
A1 = 2500
A1 A2 A3 A4
A2 = 1666
.
A4 = 1000
A3 = 1428
A1 A2
= 833 238
A1 A3
A4 A2
= 119 = 47
= 357 = 333
A1 A4
A3 A4
= 166 = 71
= 500 =142
A3 A2 =
A1 A2 A3 A2 A3 A4
HW3
1. Find the number of integers between 1 and 10,000 inclusive that are not divisible by 4, 6, 7, or 10.
Solution: Let P1 be the property that an integer is divisible by 4, P2 be the property that an integer is divisible by 6, so we also get P 3, P4 which are related to 7 and 10.Let S be the set consisting of the first 10000 positive integers. For i = 1,2,3,4 let Ai be the set consisting of those integers in S with property Pi. We wish to find the number of integers in
2 then A1 A3 = C5 =10
the same goes for other calculation.
2 =10 A3 A2 = 1 A1 A4 = C5
A4 A2 = 1 A1 A2 A3 = 0 A2 A3 A4 = 0
3 =20 A3 A4 = C6
X X X
X
X
X X
X X
we call it S R(S) =
x R(
) *R (
)
R(
) = X*R() + R(
)
R(
) = R(
)* R(
) = (2x+1) 2
R(
) = x (2 x 1) 2 = 4 x 2 5x 1
(*)
R(
) = xR (
) + R(
)Leabharlann R() = x 1
A1 A2 A3 A4
Ai A j Ak
+
=
S
-
Ai
+
Ai Ai
-
A1 A2 A3 A4
The calculation mode is similar to the problem 1, forgive me not showing it here, it is tiring...
A1 A2 A4 = 0 A1 A4 A3 = 0
A1 A2 A3 A4 = 0
S
13 = C13 41 = 560
Solution will be
A1 A2 A3 A4
=
S
-
Ai
+
Ai Ai
-
Ai A j Ak + A1 A2 A3 A4 = 96
x1 x2 x3 x4 20
that satisfy
1 x1 6
,
0 x2 7
,
4 x3 8
,
2 x4 6
Solution: We add another three variable y1, y2, y3,y4, which
y1 x1 1 , y2 x2 , y3 x3 4 , y4 x4 2 . Then we get
3. Determine the number of permutations of {1, 2, ... ,8} in which exactly four integers are in their natural positions.
Solution: First, we choose four integers that are in their natural positions, then left integers ABCD which is not in their natural position. According to the formula
S
= 10000
A1 A2 A4 A1 A4 A3
= 23
A1 A2 A3 A4
A1 A2 A3 A4
Ai A j Ak
+
=
S
-
Ai
+
Ai Ai
-
A1 A2 A3 A4
= 5429
2. Determine the number of integral solutions of the equation
A1 A2 A3 A4
(
=
13! 12! 11! 10! 9! ( ) 2!3!4!5! 3!4!5! 2!3!5! 2!3!5! 2!3!4
+
10! 9! 8! 8! 7! 6! - 7! 6! 5! 4! +3! ) ( ) 4!5! 3!5! 3!4! 2!5! 2!4! 2!3! 5! 4! 3! 2!