3 | Sequences of subsets
(a) Let n, k ≥ 1. Consider the bijection between the sequences S0 , S1 , . . . Sk of subsets
of [n] such that for any 1 ≤ i ≤ l we have either:
Si ⊃ Si−1
Si ⊂ Si−1
and
and
|Si \ Si−1 | = 1, or
|Si−1 \ Si | = 1
and the set of pairs (S0 , {bi }ki=1 ).
S0 , S1 , . . . Sk −→ (S0 , {Si 4Si−1 }ki=1 )
where Si 4Si−1 is the symmetric difference of the sets Si and Si−1 , and whose inverse
is:
(S0 , {bi }ki=1 ) −→ S0 , S1 = S0 4{b1 }, S2 = S1 4{b2 }, . . . , Sk = Sk−1 4{bk }
Now, the cardinality of the set of such pairs is 2n × nk , since there are 2n subsets of [n]
and nk k-sequences of elements in [n].
Then, the number of such sequences S0 , S1 , . . . Sk is:
2n × nk
(b) First note that the number of such sequences with the additional condition S0 =
Sk = ∅ is zero if k is odd, but this agrees with the formula:
n n n X
X
X
n
n
n
k
k
2
(n − 2i) =
(n − 2i) +
(n − 2(n − i))k
i
i
(n
−
i)
i=0
i=0 i=0
n
X
n
n
k
k
=
(n − 2i) +
(n − 2(n − i))
i
(n − i)
i=0 n
X
n
=
(n − 2i)k + (2i − n)k
i
i=0 n
X
n
=
(n − 2i)k − (n − 2i)k
i
i=0 n
X n
=
0=0
i
i=0
5
so
n 1 X n
(n − 2i)k = 0.
(n − 2i) = 0 and n
2 i=0 i
i
n X
n
i=0
k
(Idea: José Acevedo) Considering the bijection in part (a), the number of sequences
such that S0 = Sk = ∅ is just the number of k-sequences of elements in [n] such that
every element appears in an even number of positions.
k
If k = a1 +a2 +· · ·+an is an n-weak composition of k, then there are
a1 a2 . . . an
sequences of length k having exactly a1 1s, a2 2s, and so on. So what we want to count
is:
X
k
a1 a2 . . . an
a +a +···+a =k
1
2
ai ∈2Z≥0
n
Remember that:
X
k
(x1 + x2 + · · · + xn ) =
k
xa1 xa2 . . . xann
a1 a2 . . . an 1 2
a1 +a2 +···+an =k
ai ∈Z≥0
Now, we will evaluate this multinomial in all the points of the form (±1, ±1, . . . , ±1)
(2n total) and sum up.
On the left hand, there are ni ways to give values to the n variables in such a
way that exactly i terms are −1 and the remaining n − i are 1, and in this case, the
evaluation is ((−1)i + (1)(n − i))k = (n − 2i)k . Adding over all the possible values of i,
the left side is:
n X
n
(n − 2i)k
i
i=0
k
On the right hand, first fix an element
xa1 xa2 . . . xann and evalua1 a2 . . . an 1 2
ate it on all the elements (±1, ±1, . . . , ±1) and sum. Clearly, if allai ’s are even,
k
. In the case
xa11 xa22 . . . xann will always be 1, so the sum becomes: 2n
a1 a2 . . . an
that some aj is odd, then all the evaluations in (±1, . . . , ±1, 1, ±1, . . . , ±1) (where
the 1 appears in the j-th position) have the opposite sing that the evaluations in
(±1, . . . , ±1, −1, ±1, . . . , ±1), so the sum is zero. Then, adding all the evaluations
over the right hand gives:
X
X
k
k
n
n
2
=2
a1 a2 . . . an
a1 a2 . . . an
a +a +···+a =k
a +a +···+a =k
1
2
ai ∈2Z≥0
n
1
So we have the equality:
n X
n
i=0
i
k
2
ai ∈2Z≥0
X
n
(n − 2i) = 2
a1 +a2 +···+an =k
ai ∈2Z≥0
6
n
k
a1 a2 . . . an
and therefore:
n 1 X n
(n − 2i)k =
2n i=0 i
a
X
1 +a2 +···+an =k
ai ∈2Z≥0
k
a1 a2 . . . an
Thus, we have proven that the number of such sequences is:
n 1 X n
(n − 2i)k
2n i=0 i
7