Unformatted text preview: CRN: 87401 MAD 6206 Combinatorics I, Fall 2011
Assignment 2 – suggested solution
Question 1. (A recurrence for p(k, n)) Recall that p(k, n) equals the number of partitions of k into n parts. We adopt the convention that p(k, 0) = p(0, k ) = δk,0 . Also, p(k ) =
(a) Show that
p(k, n) = p(k − jn + n − 1, n − 1), k
n=0 p(k, n). n ≥ 1. j ≥1 (Hint. Consider partitions in which the smallest part(s) equals j , and then sum over j .)
(b) Find a formula for p(k, 2).
1
(c) Find a formula for p(k, 3) and show that lim k12 p(k, 3) = 12 . (Note. This one is complicated.)
k→∞ (d) Investigate the asymptotic behavior of p(k, 4).
Solution. (a) We consider the number of partitions of k into n parts so that the smallest part(s) equals
j . This equals the number of partitions of k − j into n − 1 parts so that every part is at least j . This
last number equals the number of partitions of k − j − (n − 1)(j − 1) into n − 1 parts, which equals
p(k − nj + n − 1, n − 1). Thus, we sum over j and obtain the identity.
(b) For p(k, 2), if we use the formula in (a), then
p(k, 2) = p(k − 2j + 1, 1).
j ≥1 Since p(ℓ, 1) = 1 for ℓ ≥ 1, we have
⌊k/2⌋ p(k, 2) = 1 = ⌊k/2⌋ .
j =1 Or we can argue as follows. Since the smallest part(s) is between 1 and k/2, we have
p(k, 2) = ⌊k/2⌋ .
(c) Using the formula in (a), we have
⌊k/3⌋ p(k, 3) = p(k − 3j + 2, 2) = ⌊(k − 3j + 2)/2⌋ .
j =1 j ≥1 The above sum gives that for m ≥ 1, 1 2 (3m2 + 1)/2 ,
if k = 3m, 1
p(k, 3) = 2 ⌈(3m − 1)(m + 1)/2⌉ , if k = 3m + 1, p(3m, 3) + m,
if k = 3m + 1. To illustrate how to manipulate the ﬂoor function, we consider the sum
m m−1 p(3m, 3) = ⌊(3m − 3j + 2)/2⌋ =
j =1 ⌊(3j + 2)/2⌋
j =0 m−1 (3j + 2)/2 − 1 {j : j is odd, 0 ≤ j ≤ m − 1}
2 =
j =0 = m + 3 m(m − 1) − 1 ⌈(m − 1)/2⌉
4
2
=
= 1
2
1
2 2m + 3 m(m − 1) − (m − 1)/2
2
(3m2 + 1)/2 . The above formula gives that
p(k, 3) = 12
12 k + O(k ), as k → ∞. Notes on the bigO notation. Two functions f, g have the property f (k ) = O(g (k )), as k → ∞, if there is a
constant C so that f (k )/g (k ) ≤ C for all suﬃciently large k . Thus, the above formula for p(k, 3) means
that there are constants C and k0 so that
p(k, 3) − 12
12 k ≤ Ck, ∀k ≥ k0 . If we really want to have a “single” formula for p(k, 3):
p(k, 3) = 1
4 ⌊k/3⌋ (2k − 3 ⌊k/3⌋ + 1) − 1 (⌊k/6⌋ + ⌊(k + 2)/6⌋ − ⌊(k + 1)/6⌋),
2
⌊k/3⌋
j =1 ⌊(k we can proceed as follows. We note ﬁrst that p(k, 3) =
k − 3j + 2 ≡ k − j − 3j + 2)/2⌋, and that (mod 2). Thus,
⌊k/3⌋
1
2 (k p(k, 3) = − 3j + 2 − ǫk,j ), j =1 where ǫk,j = 0 if k − j is even, and ǫk,j = 1 if k − j is odd. Hence we have
⌊k/3⌋ p(k, 3) = 1
4 ⌊k/3⌋ (2k − 6j + 4) −
j =1 1
2 ⌊k/3⌋ ǫk,j = 1
4 ⌊k/3⌋ (2k − 3 ⌊k/3⌋ + 1) − 1
2 j =1 ǫk,j .
j =1 It therefore remains to show that
⌊k/3⌋ ǫk,j = ⌊k/6⌋ + ⌊(k + 2)/6⌋ − ⌊(k + 1)/6⌋ .
j =1 To do this, we note that if k is odd, then
⌊k/3⌋ ǫk,1 = 0, ǫk,2 = 1, ǫk,3 = 0, . . . , ǫk,j = 1
2 ⌊k/3⌋ , ǫk,j = and 1
2 ⌊k/3⌋ , j =1 and when k is even, we have
⌊k/3⌋ ǫk,1 = 1, ǫk,2 = 0, ǫk,3 = 1, . . . , and
j =1 1
1
Now 2 ⌊k/3⌋ , when k is odd, and 2 ⌊k/3⌋ , when k is even, are almost the same. They are both equal
1
to 2 ⌊k/3⌋ = ⌊k/6⌋ except when k is even and ⌊k/3⌋ is odd, which is the same as k ≡ 4 (mod 6). Hence,
⌊k/3⌋ ǫk,j = ⌊k/6⌋ + δk ,
j =1 where δk = 1 if k ≡ 4 (mod 6) and δk = 0 if otherwise. Since
δk = ⌊(k + 2)/6⌋ − ⌊(k + 1)/6⌋ ,
we now have
⌊k/3⌋ ǫk,j = ⌊k/6⌋ + ⌊(k + 2)/6⌋ − ⌊(k + 1)/6⌋ .
j =1 (d) For p(k, 4), we have from (c) that p(k, 3) = 12
12 k + O(k ). Thus, ⌊k/4⌋
1
12 (k p(k − 4j + 3, 3) = p(k, 4) = − 4j + 3)2 + O(k 2 ) j =1 j ≥1 ⌊k/4⌋ =
= (k 2 − 8kj + 16j 2 ) + O(k 2 ) 1
12
k3 j =1 = 1
1
12
12 ( 4 − 8 · 2 · ( 4 )
2
13
144 k + O (k ). + 16 · 1
3 · ( 1 )3 ) + O(k 2 )
4 Question 2. (Diagonal elements of p(k, n)) We know that p(k, k ) = 1 and p(k, k + i) = 0 for i ≥ 1. How
about p(k, k − i)?
(a) We proved in class that p(k, n) = n p(k − n, i). Use this to prove that for k ≥ 2n ≥ 0,
i=0
p(k, k − n) = p(n).
(b) Find formulas for p(k, k − 1), p(k, k − 2), p(k, k − 3).
Solution. (a) When k ≥ 2n ≥ 0, we have i=0 i=0 p(n, i) = p(n), p(n, i) = p(k − (k − n), i) = p(k, k − n) = n k −n k −n i=0 where the last but one inequality follows from k − n ≥ n and p(n, i) = 0 for i > n.
(b) We use the formula in (a) to get
p(k, k − 1) = p(1) = 1,
p(k, k − 2) = p(2) = 2,
p(k, k − 3) = p(3) = 3, ∀k ≥ 2,
∀k ≥ 4,
∀k ≥ 6, p(1, 0) = 0,
p(3, 1) = 1, p(2, 0) = 0,
p(5, 2) = p(3, 0) + p(3, 1) + p(3, 2) = 2, p(4, 1) = 1, p(3, 0) = 0. Question 3. (Stirling numbers and power sums) (a) We showed in class that k
1 = 1, k
2 = 2k−1 − 1, k
k
k
and k = 1 (3k − 3 · 2k + 3). We also know that k = 1. Find formulas for k−1 , k−2 and k−3 .
3
6
k
(b) Let ∆f (x) = f (x + 1) − f (x). Prove that for k ≥ 1, ∆(x)k = k (x)k−1 . That is, written in another form
where k is increased by 1, (x)k = 1
((x + 1)k+1 − (x)k+1 ) ,
k+1 k ≥ 0. Prove that for integers n ≥ m ≥ 0 and k ≥ 0,
n (x + i)k =
i=m 1
((x + n + 1)k+1 − (x + m)k+1 ).
k+1 (c) Prove the following formula for power sums:
m n km =
j =0 k=0 m (n + 1)j +1
.
j
j+1 (d) We have standard formulas for m = 1, 2, 3. Use the formula in (c) to ﬁnd a formula for the power sum
when m = 4.
k
Solution. (a) For k−1 , we note that each set partition counted is uniquely determined the block containing
two elements. Thus,
k
k
=
,
k ≥ 1.
k−1
2
k
For k−2 , we note that each set partition counted either contains a block with three elements or two
blocks with 2 elements each. Thus, k
k−2
For k
k −3 k
1k
+
22
3 = k−2
2 = 1
k (k − 1)(k − 2)(3k − 5).
24 , we know that most blocks are singleton sets except that we can have
k
1
3! 2,2,2,k−6 (i) three blocks with 2 elements each: partitions, or (ii) one block with 3 elements and 1 block with 2 elements:
k
4,k−4 (iii) one block with 4 elements:
Thus,
k
k−3 k
3,2,k−5 partitions. k
k
k
+
+
2, 2, 2, k − 6
3, 2, k − 5
4 = 1
3! = 1
48 k (k partitions, or = 1
48 (k )6 + 1
12 (k )5 + 1
24 (k )4 − 1)(k − 2)2 (k − 3)2 . (b) We have
∆(x)k = (x + 1)k − (x)k = (x + 1)(x)k−1 − (x)k−1 (x − k + 1) = k (x)k−1 .
Thus,
n n (x + i)k =
i=m j =m 1
1
((x + i + 1)k+1 − (x + i)k+1 ) =
((x + n + 1)k+1 − (x + m)k+1 ).
k+1
k+1 (c) By expanding k m as a sum of falling factorials using Stirling numbers of the second kind, we have
n n m km =
k=0 k=0 j =0 m
(k )j =
j m
j =0 m
j n (k )j .
k=0 Using the result in (b) with x = m = 0, we have for j ≥ 0,
n (k )j =
k=0 We therefore obtain (n + 1)j +1
1
((n + 1)j +1 − (0)j +1 ) =
.
j+1
j+1
m n k
k=0 m =
j =0 m (n + 1)j +1
.
j
j+1 (d) When m = 4, we have
n km = 1
2 k=0 =
= 4
(n + 1)2 +
1 1
3 4
(n + 1)3 +
2 4
(n + 1)4 +
3 1
4 1
7
6
1
2 (n + 1)2 + 3 (n + 1)3 + 4 (n + 1)4 + 5 (n
2
1
30 n(n + 1)(2n + 1)(3n + 3n − 1). 1
5 4
(n + 1)5
4 + 1)5 Question 4. (Stirling numbers of the ﬁrst kind) We deﬁned in class the falling factorial (x)n and the
rising factorial (x)n as
n−1 n−1 (x − i), (x)n = (x)n = (x + i),
i=0 i=0 and we deﬁned the Stirling number of the ﬁrst kind, s(k, n), and the unsigned Stirling number of the ﬁrst
k
kind, n , by
k
s(k, n) = [xn ](x)k ,
= [xn ](x)k .
n
k
We showed in class that n = (−1)k−n s(k, n) = s(k, n). We shall show some additional properties of
in this problem.
(a) Find for integer k ≥ 0,
k k k
,
n n=0 k
(−1)n .
n and
n=0 (b) Prove that for integer k ≥ 0,
k n
n=0 k n−1
x
= (x)k
n k −1
i=0 1
.
x+i Rewrite the formula for x = 0, −1, . . . , −k + 1.
(c) Find for integer k ≥ 0,
k
n=0
n even k
n k n and
n=0
n even k
.
n Solution. (a) From the identity
k (x)k =
n=0 kn
x,
n integer k ≥ 0, we obtain, by setting x = 1, that
k
n=0 k
= (1)k = k !,
n integer k ≥ 0. and by setting x = −1 that
k
n=0 1, k
k
n
(−1) = (−1) = −1, n 0, k = 0,
k = 1,
k ≥ 2. k
n (b) Diﬀerentiating with respect to x both sides of the identity
k kn
x,
n (x)k =
n=0 integer k ≥ 0, we obtain
n−1
k (x) i=0 1
=
x+i k n
n=0 k n−1
x
.
n If x = −m, where m = 0, 1, . . . , n − 1, then the formula is simply
k
n=0 k
n
(−m)n−1 = lim (z )k
z →−m
n k −1
i=0 1
z+i (c) Notice that using the formulas found in (a), we have
k
n=0
n even k k
1
=
2
n n=0 k
+
n k
n=0 0, = 1, (−1)m m!(k − m − 1)!, 1, = 0,
1 2 k !, k
(−1)n
n k = 0,
k = 1,
k ≥ 2. k = 0,
k = 1,
k ≥ 2. and using the formula in (b), we get (by setting x = 1 and x = −1) that
k n
n=0 where Hk = k
1
j =1 j k
= k!
n n=0 k
n
(−1)n−1
n Then
n
n=0
n even i=0 1
= k !H k ,
1+i is the k th harmonic number, and
k n k −1 k
1
=
2
n n n
n=0 k
+
n k n
n=0 0, = 1, −(k − 2)!, k
(−1)n
n k = 0,
k = 1,
k ≥ 2. 0, = 1,
1 2 (k !Hk + (k − 2)!), k = 0,
k = 1,
k ≥ 2. Note that the quantities in this problem have combinatorial meaning. We shall see later that the number
k
of permutations on [k ] with n cycles equals n . Then, for example, the average number of cycles in a
permutation on [n] equals
1
k! k n
n=0 k
= Hk ∼ ln(k ),
n as k → ∞. Question 5. (Inclusionexclusion) Let h, k be nonnegative integers and n ∈ Z+ . Find the number of
integer solutions to the following system:
x1 + x2 + · · · + xn = k, where 0 ≤ xi ≤ h, i = 1, 2, . . . , n. What should your answer be if h ≥ k ? For what values of h and k should your answer be 0?
Solution. Let Ai be the set of integer solutions to x1 + x2 + · · · + xn = k subject to the conditions that
xj ≥ 0 for all j and xi > h. Also, let N (m) be the number of integer solutions to
x1 + x2 + · · · + xn = m, xi ≥ 0. Then
N (m) = = 1
n−1 ,
(n−1)! (m) 0, m ≥ 0,
otherwise. n
j
j =0 (−1) Sj , Then by PIE, the answer equals
Sj = m+n−1
n−1 where n
A1 ∩ A2 ∩ · · · ∩ Aj  =
j n
N (k − j (h + 1)).
j Thus, the answer equals
n (−1)j
j =0 n
N (k − j (h + 1)).
j n
If h ≥ k , then the upper bound h on xi is not needed and the answer is k+−−1 . The answer equals 0
n1
when nh < k . For example, if k = nh + 1 + m, where m is a nonnegative integer, then N (k − j (h + 1)) = N ((n − j )h + 1 + m − j ) =
and we have n (−1)j
j =0 n
j (n + j )(h + 1) + m
n−1 = 0, (n + j )(h + 1) + m
,
n−1 integers h, m ≥ 0, which is a rather strange identity.
Question 6. (Inclusionexclusion) After playing in the school playground in a winter day, m children
take oﬀ their wet mittens and leave them on a bench to dry. (We shall assume that each child has two
mittens, and all 2m mittens are diﬀerent.) Each child then takes two mittens at the end of the day. Find
the number of distributions of the mittens to the children under the following situations:
(a) Each child gets any two mittens, and every child gets at least one mitten belonging to someone else.
(b) Each child gets a mitten for the left hand and a mitten for the right hand, and every child gets at least
one mitten belonging to someone else.
(c) Each child gets a mitten for the left hand and a mitten for the right hand, and every child gets both
mittens belonging to someone else.
Solution. (a) The objects are distributions of mittens to children so that each child has two. Let Ai be
the set of distributions in which child i gets his/her own mittens. Now, using PIE, the answer equals
m
j
j =0 (−1) Sj . To calculate Sj , we note ﬁrst that each term in Sj are equal because of symmetry, and since
there are m
j terms in Sj , we have
Sj = m
A1 ∩ A2 ∩ . . . ∩ Aj .
j But
A1 ∩ A2 ∩ . . . ∩ Aj  = (2m − 2j )!
.
2m−j Hence, answer equals
m (−1)j
j =0 m (2m − 2j )!
.
j
2m−j (b) The objects are distributions of mittens so that each child gets a left mitten and a right mitten. Let Ai
be the set of distributions in which child i gets his/her own mittens. Then using PIE, the answer equals
m m (−1)j Sj (−1)j j =0 m
A1 ∩ A2 ∩ . . . ∩ Aj 
j (−1)j = m
(m − j )!2 .
j j =0
m =
j =0 (c) Each distribution counted in this case is made of a derangement on the left mittens and a derangement
on the right mittens. Hence the answer equals 2
Dm = m! m
j =0 (−1)j
j! 2 . k
n We give below tables of values for k
0
1
2
3
4
5
6
7
8 p(k, 0)
1
0
0
0
0
0
0
0
0 and p(k, n). You can ﬁnd these values in most text books. k
0 k
1 k
2 k
3 k
4 k
5 k
6 k
7 k
8 1
0
0
0
0
0
0
0
0 k
0
1
2
3
4
5
6
7
8 k
0
1
2
3
4
5
6
7
8 k
n , 1
1
1
1
1
1
1
1 1
3
7
15
31
63
127 1
6
25
90
301
966 1
10
65
350
1701 1
15
140
1050 1
21
266 1
28 1 k
0 k
1 k
2 k
3 k
4 k
5 k
6 k
7 k
8 1
0
0
0
0
0
0
0
0 1
1
2
6
24
120
720
5040 1
3
11
50
274
1764
13068 1
6
35
225
1624
13132 1
10
85
735
6769 1
15
175
1960 1
21
322 1
28 1 p(k, 1) p(k, 2) p(k, 3) p(k, 4) p(k, 5) p(k, 6) p(k, 7) p(k, 8) 1
1
1
1
1
1
1
1 1
1
2
2
3
3
4 1
1
2
3
4
5 1
1
2
3
5 1
1
2
3 1
1
2 1
1 1 ...
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 n=0, sj

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