American Mathematical Monthly Problem 11424 by Emeric Deutsch (edited)
For every k ∈ N, for every k-tuple v ∈ {0,1}k and every i ∈ {1,2, ..., k}, let us denote by vi the i-th component of the k-tuple v (remember that v is an element of {0,1}k, that is, a k-tuple of elements of {0,1}). Then, every v ∈ {0,1}k satisfies v = (v1, v2, ..., vk).
Let n >1 be an integer. For any n-tuple a∈ {0,1}n, we define two integers F (a) and G(a) by
F (a) =|{i∈ {1,2, ..., n−1} |ai =ai+1 = 0}|; G(a) =|{i∈ {1,2, ..., n−1} |ai =ai+1 = 1}|. Set
T ={a ∈ {0,1}n|F (a) =G(a)}. Prove that
|T|=
2
n−2 (n−2)2
, if n is even;
n−1 (n−1)2
, if n is odd
= 2
n−2 b(n−2)2c
.
Example. For instance, if n = 4, then
T ={(0,0,1,1),(0,1,0,1),(1,0,1,0),(1,1,0,0)}, so that |T|= 4.
Solution by Darij Grinberg.
First, we are going to introduce some notations:
• For any assertion U, we denote by [U] the Boolean value of the assertion U (that is, [U] =
1, ifU is true;
0, ifU is false ).
It is then clear that ifB is a set, and U(a) is an assertion for every element a of B, then
X
a∈B
[U(a)] =|{a∈B | U(a) is true}|.
1
Also, if U1, U2, ..., Um are m assertions, then [U1 and U2 and ...and Um] =
m
Q
j=1
[Uj].
1In fact, X
a∈B
[U(a)] =X
a∈B
1, if U(a) is true;
0, if U(a) is false = X
a∈B such that U(a) is true
1 + X
a∈B such that U(a) is false
0
| {z }
=0
= X
a∈B such that U(a) is true
1
=|{a∈B | U(a) is true}|.
• Ifn is an integer andk is a real, then we define the binomial coefficient n
k
by n
k
=
( n(n−1)...(n−k+ 1)
k! , if k∈N;
0, otherwise
,
where N denotes the set {0,1,2, ...}. This definition agrees with the standard definition of
n k
in the case whenk ∈Z.
A consequence of this definition is that if S is a finite set andk is a real, then |S|
k
=|{A∈ P(S)| |A|=k}|
2.
A simple fact:
Lemma 1. Every integerη >0 satisfies 2η
η
= 2
2η−1 η−1
.
3
Notice that
[α= 1] =α for every α ∈ {0,1}, (1)
since
α=
1, if α = 1;
0, if α6= 1 (since α∈ {0,1})
=
1, if α = 1 is true;
0, if α= 1 is false = [α = 1].
2In fact, ifk∈N, then this is well-known, and otherwise |S|
k
=|{A∈ P(S)| |A|=k}|follows
from |S|
k
= 0 and
A∈ P(S)| |A|=k
| {z }
impossible, since k6∈N
=|∅|= 0
3Proof. We have 2η
η
= (2η)!
η!·(2η−η)! = 2η·(2η−1)!
(η·(η−1)!)·(2η−η)! (since (2η)! = 2η·(2η−1)! andη! =η·(η−1)!)
= 2η η
|{z}
=2
· (2η−1)!
(η−1)!·(2η−η)!
| {z }
= (2η−1)!
(η−1)!·((2η−1)−(η−1))!
(since 2η−η=(2η−1)−(η−1))
= 2· (2η−1)!
(η−1)!·((2η−1)−(η−1))!
| {z }
=
2η−1 η−1
= 2
2η−1 η−1
, qed.
Also,
[α= 0] = 1−α for every α∈ {0,1}, (2) since
α=
0, if α = 0;
1, if α6= 0 (since α∈ {0,1})
=
1−1, if α= 0;
1−0, if α6= 0 = 1−
1, if α= 0;
0, if α6= 0 = 1−
1, if α= 0 is true;
0, if α = 0 is false; = 1−[α= 0]. Now, let us solve the problem.
Define a map f :{0,1}n → P({1,2, ..., n}) by
f(a) = {i∈ {1,2, ..., n} |ai = 1} for any n-tuplea ∈ {0,1}n. Define a map g :P({1,2, ..., n})→ {0,1}n by
g(A) = ([1∈A],[2∈A], ...,[n ∈A]) for any A∈ P({1,2, ..., n}). Then, for any n-tuple a∈ {0,1}n, we have
(g◦f) (a) = g(f(a)) = ([1∈f(a)],[2∈f(a)], ...,[n ∈f(a)])
= ([a1 = 1],[a2 = 1], ...,[an= 1])
(since [i∈f(a)] = [ai = 1] for any i∈ {1,2, ..., n} by the definition of f)
= (a1, a2, ..., an) (since [ai = 1] =ai for any i∈ {1,2, ..., n} by (1), since ai ∈ {0,1})
=a.
Thus, g ◦f = id. On the other hand, for any A ∈ P({1,2, ..., n}), we have A ⊆ {1,2, ..., n} and thus
(f◦g) (A) = f(g(A)) ={i∈ {1,2, ..., n} |(g(A))i = 1}
={i∈ {1,2, ..., n} |[i∈A] = 1}
since g(A) = ([1∈A],[2∈A], ...,[n∈A]) and thus (g(A))i = [i∈A] for every i∈ {1,2, ..., n}
={i∈ {1,2, ..., n} |i∈A}=A∩ {1,2, ..., n}=A (since A⊆ {1,2, ..., n}). Therefore,f ◦g = id.
Since g◦f = id and f◦g = id, the mapsf and g are mutually inverse. Hence, the map f :{0,1}n→ P({1,2, ..., n}) is a bijection.
For any n-tuple a∈ {0,1}n, we have
|f(a)|=|{i∈ {1,2, ..., n} |ai = 1}|= X
i∈{1,2,...,n}
[ai = 1]
| {z }
=aiby (1), sinceai∈{0,1}
= X
i∈{1,2,...,n}
ai.
Thus, for any realk, we have
a∈ {0,1}n| X
i∈{1,2,...,n}
ai =k
=|{a∈ {0,1}n | |f(a)|=k}|
=
f−1({A∈ P({1,2, ..., n})| |A|=k})
=|{A∈ P({1,2, ..., n})| |A|=k}|
(since f is a bijection)
=
|{1,2, ..., n}|
k
= n
k
.
Applying this to n−2 instead of n, we obtain
a ∈ {0,1}n−2 | X
i∈{1,2,...,n−2}
ai =k
=
n−2 k
. (3)
Now, for any n-tuple a∈ {0,1}n, we have
F (a) =|{i∈ {1,2, ..., n−1} |ai =ai+1 = 0}|= X
i∈{1,2,...,n−1}
[ai =ai+1 = 0]
= X
i∈{1,2,...,n−1}
[ai = 0 andai+1 = 0] = X
i∈{1,2,...,n−1}
[ai = 0] [ai+1 = 0] = X
i∈{1,2,...,n−1}
(1−ai) (1−ai+1) since [ai = 0] = 1−ai (by (2), applied to α=ai ∈ {0,1})
and [ai+1 = 0] = 1−ai+1 (by (2), applied to α=ai+1 ∈ {0,1})
; G(a) =|{i∈ {1,2, ..., n−1} |ai =ai+1 = 1}|= X
i∈{1,2,...,n−1}
[ai =ai+1 = 1]
= X
i∈{1,2,...,n−1}
[ai = 1 andai+1 = 1] = X
i∈{1,2,...,n−1}
[ai = 1] [ai+1 = 1] = X
i∈{1,2,...,n−1}
aiai+1 since [ai = 1] =ai (by (1), applied to α =ai ∈ {0,1})
and [ai+1 = 1] =ai+1 (by (1), applied to α=ai+1 ∈ {0,1})
,
so that
F (a)−G(a) = X
i∈{1,2,...,n−1}
(1−ai) (1−ai+1)− X
i∈{1,2,...,n−1}
aiai+1
= X
i∈{1,2,...,n−1}
(1−ai) (1−ai+1)
| {z }
=1−ai−ai+1+aiai+1
−aiai+1
= X
i∈{1,2,...,n−1}
(1−ai −ai+1)
= X
i∈{1,2,...,n−1}
1
| {z }
=(n−1)·1=n−1
− X
i∈{1,2,...,n−1}
ai
| {z }
=a1+ P
i∈{2,3,...,n−1}
ai
− X
i∈{1,2,...,n−1}
ai+1
| {z }
= P
i∈{2,3,...,n}
ai
= P
i∈{2,3,...,n−1}
ai+an
= (n−1)−
a1+ X
i∈{2,3,...,n−1}
ai
−
X
i∈{2,3,...,n−1}
ai+an
= (n−1)−a1−an−2 X
i∈{2,3,...,n−1}
ai. (4)
Define a map u:{0,1}n→ {0,1}n−2 by
u(a) = (a2, a3, ..., an−1) for any n-tuplea ∈ {0,1}n. Then, (u(a))i =ai+1 for any n-tuplea ∈ {0,1}n and any i∈ {1,2, ..., n−2}.
Define a map v :{0,1}n→ {0,1} × {0,1}n−2× {0,1}by
v(a) = (a1, u(a), an) for any n-tuple a∈ {0,1}n.
Define a map w:{0,1} × {0,1}n−2× {0,1} → {0,1}n by
w(x, b, y) = (x, b1, b2, ..., bn−2, y) for any triple (x, b, y)∈ {0,1}×{0,1}n−2×{0,1}. Then, any n-tuple a∈ {0,1}n satisfies
(u(a))i =ai+1 for any i∈ {1,2, ..., n−2}, (5) and thus
(w◦v) (a) =w(v(a)) =w(a1, u(a), an)
=
a1, (u(a))1
| {z }
=a1+1by (5)
, (u(a))2
| {z }
=a2+1 by (5)
, ..., (u(a))n−2
| {z }
=a(n−2)+1by (5)
, an
= a1, a1+1, a2+1, ..., a(n−2)+1, an
= (a1, a2, a3, ..., an−1, an) =a, so that w◦v = id. Besides, any triple (x, b, y)∈ {0,1} × {0,1}n−2× {0,1} satisfies
w(x, b, y) = (x, b1, b2, ..., bn−2, y), so that
(w(x, b, y))i+1 =bi for any i∈ {1,2, ..., n−2}, (6) and thus
u(w(x, b, y)) = (w(x, b, y))2,(w(x, b, y))3, ...,(w(x, b, y))n−1
=
(w(x, b, y))1+1
| {z }
=b1 by (6)
,(w(x, b, y))2+1
| {z }
=b2 by (6)
, ...,(w(x, b, y))(n−2)+1
| {z }
=bn−2 by (6)
= (b1, b2, ..., bn−2) =b, so that
(v◦w) (x, b, y) =v(w(x, b, y)) = ((w(x, b, y))1, u(w(x, b, y)),(w(x, b, y))n)
= (x, b, y)
since (w(x, b, y))1 = (x, b1, b2, ..., bn−2, y)1 =x,
u(w(x, b, y)) =b and (w(x, b, y))n = (x, b1, b2, ..., bn−2, y)n=y
, and thus v◦w= id.
Since v ◦w = id and w◦v = id, the maps v and w are mutually inverse. Hence, the map v is a bijection.
Now, for any x∈ {0,1} and y ∈ {0,1}, we have {a∈T |(a1, an) = (x, y)}
=T ∩ {a∈ {0,1}n|(a1, an) = (x, y)}
={a∈ {0,1}n|F (a) = G(a)} ∩ {a ∈ {0,1}n|(a1, an) = (x, y)}
={a∈ {0,1}n|F (a) = G(a) and (a1, an) = (x, y)}
={a∈ {0,1}n|F (a) = G(a) anda1 =x and an=y}
={a∈ {0,1}n|F (a)−G(a) = 0 and a1 =x and an =y}
=
a∈ {0,1}n|(n−1)−a1−an−2 X
i∈{2,3,...,n−1}
ai = 0 and a1 =x and an=y
(by (4))
=
a∈ {0,1}n|2 X
i∈{2,3,...,n−1}
ai = (n−1)−a1−an and a1 =x and an =y
=
a∈ {0,1}n|2 X
i∈{2,3,...,n−1}
ai = (n−1)−x−y and a1 =x and an =y
=
a∈ {0,1}n|2 X
i∈{2,3,...,n−1}
(u(a))i−1 = (n−1)−x−y and a1 =x and an=y
because u(a) = (a2, a3, ..., an−1), so that ai = (u(a))i−1 for any i∈ {2,3, ..., n−1}
=
a∈ {0,1}n|2 X
i∈{1,2,...,n−2}
(u(a))i = (n−1)−x−y and a1 =x and an=y
since X
i∈{2,3,...,n−1}
(u(a))i−1 = X
i∈{1,2,...,n−2}
(u(a))i
=
a∈ {0,1}n| X
i∈{1,2,...,n−2}
(u(a))i = ((n−1)−x−y)2 and a1 =x and an =y
=
a∈ {0,1}n|u(a)∈
b ∈ {0,1}n−2 | X
i∈{1,2,...,n−2}
bi = ((n−1)−x−y)2
and a1 =x and an =y
since P
i∈{1,2,...,n−2}
(u(a))i = ((n−1)−x−y)2 is equivalent to u(a)∈
(
b∈ {0,1}n−2 | P
i∈{1,2,...,n−2}
bi = ((n−1)−x−y)2 )
=
a∈ {0,1}n|u(a)∈
b ∈ {0,1}n−2 | X
i∈{1,2,...,n−2}
bi = ((n−1)−x−y)2
and a1 ∈ {x} and an ∈ {y}
=
a∈ {0,1}n|(a1, u(a), an)∈ {x} ×
b ∈ {0,1}n−2 | X
i∈{1,2,...,n−2}
bi = ((n−1)−x−y)2
× {y}
=
a∈ {0,1}n|v(a)∈ {x} ×
b ∈ {0,1}n−2 | X
i∈{1,2,...,n−2}
bi = ((n−1)−x−y)2
× {y}
=v−1
{x} ×
b ∈ {0,1}n−2 | X
i∈{1,2,...,n−2}
bi = ((n−1)−x−y)2
× {y}
. Since v is a bijection, this yields
|{a∈T |(a1, an) = (x, y)}|
=
{x} ×
b ∈ {0,1}n−2 | X
i∈{1,2,...,n−2}
bi = ((n−1)−x−y)2
× {y}
=|{x}|
| {z }
=1
·
b ∈ {0,1}n−2 | X
i∈{1,2,...,n−2}
bi = ((n−1)−x−y)2
· |{y}|
| {z }
=1
=
b ∈ {0,1}n−2 | X
i∈{1,2,...,n−2}
bi = ((n−1)−x−y)2
=
a ∈ {0,1}n−2 | X
i∈{1,2,...,n−2}
ai = ((n−1)−x−y)2
=
n−2
((n−1)−x−y)2
(by (3), applied to k = ((n−1)−x−y)2). (7)
Now,
T ={a∈T}={a∈T |(a1, an)∈ {(0,0),(0,1),(1,0),(1,1)}}
because any a∈T satisfiesa ∈ {0,1}n, thus a1 ∈ {0,1} and an∈ {0,1}, and hence (a1, an)∈ {0,1} × {0,1}={(0,0),(0,1),(1,0),(1,1)}
={a∈T |(a1, an) = (0,0) or (a1, an) = (0,1) or (a1, an) = (1,0) or (a1, an) = (1,1)}
={a∈T |(a1, an) = (0,0)} ∪ {a∈T |(a1, an) = (0,1)}
∪ {a∈T |(a1, an) = (1,0)} ∪ {a ∈T |(a1, an) = (1,1)}. Since the four sets
{a ∈T |(a1, an) = (0,0)}, {a∈T |(a1, an) = (0,1)}, {a ∈T |(a1, an) = (1,0)}, {a∈T |(a1, an) = (1,1)}
are pairwise disjoint (because every element a of T lies in at most one of these four sets, since every elementa of T satisfies at most one of the four equations
(a1, an) = (0,0), (a1, an) = (0,1), (a1, an) = (1,0), (a1, an) = (1,1) ), this yields
|T|=|{a∈T |(a1, an) = (0,0)}|
| {z }
=
n−2
((n−1)−0−0)2
(by (7), applied tox=0 andy=0)
+|{a∈T |(a1, an) = (0,1)}|
| {z }
=
n−2
((n−1)−0−1)2
(by (7), applied tox=0 andy=1)
+|{a∈T |(a1, an) = (1,0)}|
| {z }
=
n−2
((n−1)−1−0)2
(by (7), applied tox=1 andy=0)
+|{a∈T |(a1, an) = (1,1)}|
| {z }
=
n−2
((n−1)−1−1)2
(by (7), applied tox=1 andy=1)
=
n−2
((n−1)−0−0)2
+
n−2
((n−1)−0−1)2
+
n−2
((n−1)−1−0)2
+
n−2
((n−1)−1−1)2
=
n−2 (n−1)2
+
n−2 (n−2)2
+
n−2 (n−2)2
+
n−2 (n−3)2
=
n−2 (n−1)2
+ 2
n−2 (n−2)2
+
n−2 (n−3)2
. (8)
Now, ifnis odd, thenn−2 is odd, so that (n−2)2∈/ N,and thus
n−2 (n−2)2
=
0,so that
|T|=
n−2 (n−1)2
+ 2
n−2 (n−2)2
| {z }
=0
+
n−2 (n−3)2
(by (8))
=
n−2 (n−1)2
+
n−2 (n−3)2
=
n−2 (n−1)2
+
n−2 (n−1)2−1
=
(n−2) + 1 (n−1)2
by the recurrence of the binomial coefficients, since (n−1)2∈Z (since n−1 is even, since n is odd)
and (n−1)2>0 (since n >1)
=
n−1 (n−1)2
(9)
=
2 (n−1)2 (n−1)2
(sincen−1 = 2 (n−1)2)
= 2
2 (n−1)2−1 (n−1)2−1
by Lemma 1, applied toη= (n−1)2, since (n−1)2 is an integer and (n−1)2>0
= 2
n−2 (n−3)2
since 2 (n−1)2−1 = (n−1)−1 =n−2 and (n−1)2−1 = (n−3)2
= 2
n−2 b(n−2)2c
(10) (since (n−3)2 =b(n−2)2c 4). On the other hand, ifn is even, thenn−1 and n−3 are odd, so that (n−1)2∈/ N and (n−3)2∈/ N, and thus
|T|=
n−2 (n−1)2
| {z }
=0,since (n−1)2∈/N
+2
n−2 (n−2)2
+
n−2 (n−3)2
| {z }
=0,since (n−3)2∈/N
(by (8))
= 2
n−2 (n−2)2
(11)
= 2
n−2 b(n−2)2c
(12) (since (n−2)2 = n2
| {z }
∈Z(sincenis even)
−1∈Z yields (n−2)2 =b(n−2)2c).
Combining (11) and (9), we obtain
|T|=
2
n−2 (n−2)2
, if n is even;
n−1 (n−1)2
, if n is odd
4This is because (n−3)2 = (n−1)2
| {z }
∈Z
−1 ∈ Z and (n−2)2 = (n−3)2 + 12 ∈ [(n−3)2,(n−3)2 + 1[.
for any n >1. On the other hand, combining (12) and (10), we obtain
|T|=
2
n−2 b(n−2)2c
, if n is even;
2
n−2 b(n−2)2c
, if n is odd
= 2
n−2 b(n−2)2c
for any n >1. This solves the problem.
