Constants

## 2852683

https://math.stackexchange.com/questions/2852683

How many subsets of size k from {1,2,…n} such that if subset contains 2 it doesn't contain 1

### **answer 1**

(

n
k

)

−

(

n
−
2

k
−
2

)

.

{\displaystyle {\binom {n}{k}}-{\binom {n-2}{k-2}}.}

"

{
1
,
2
}
⊆
A

{\displaystyle \{1,2\}\subseteq A}

" is false.

### **answer 2**

"does not contain

1

{\displaystyle 1}

 if it contains $2$" is the same as "does not contain $1$ or does not contain $2$".

Let $\mathcal S\_1$ $\mathcal S\_1$ denote the subcollection of subsets of $\{1,\dots,n\}$ having size k and such that $i\notin S$ for every

S
∈

S

i

{\displaystyle S\in {\mathcal {S}}\_{i}}

.

So actually

S

i

:=
{
S
∈
℘
(
{
1
,
…
,
n
}
)
∣
i
∉
S
∧

S

=
k
}

{\displaystyle {\mathcal {S}}\_{i}:=\{S\in \wp (\{1,\dots ,n\})\mid i\notin S\wedge |S|=k\}}

Then with inclusion/exclusion and symmetry we find:

S

1

∪

S

2

=

S

1

S

2

−

S

1

∩

S

2

=
2

S

1

−

S

1

∩

S

2

=
2

(

n
−
1

k

)

−

(

n
−
2

k

)

### **answer 3**

Let

E
(
A
,
X
)

=

d
e
f

A
.

E
′

(
X
)

{\displaystyle E(A,X)=\_{def}A.E'(X)}

 the species of sets of sort X that contains also one A.

then the species in the problem is:

E

k

E

m

(
A
,
B
,
X
# )
A
.
B
.
(

E

k

E

m

)
″

(
X
# )

{\displaystyle E\_{k}E\_{m}(A,B,X)=A.B.(E\_{k}E\_{m})''(X)=}

E

k

(
A
,
B
,
X
)
.

E

m

(
X
)

E

k

(
A
,
X
)
.

E

m

(
B
,
X
)

E

k

(
B
,
X
)
.

E

m

(
A
,
X
)

E

k

(
X
)
.

E

m

(
A
,
B
,
X
)

{\displaystyle E\_{k}(A,B,X).E\_{m}(X)+E\_{k}(A,X).E\_{m}(B,X)+E\_{k}(B,X).E\_{m}(A,X)+E\_{k}(X).E\_{m}(A,B,X)}

The coefficient

x

n
−
2

(
n
−
2
)
!

{\displaystyle {\frac {x^{n-2}}{(n-2)!}}}

 of the sum of the last three terms is the required number.

(

n
−
2

k
−
1

)

(

n
−
2

k
−
1

)

(

n
−
2

k

)

{\displaystyle {\binom {n-2}{k-1}}+{\binom {n-2}{k-1}}+{\binom {n-2}{k}}}

Why do we have

(

n
k

)

{\displaystyle \binom{n}{k}}

 and

(

n
−
1

k

)

{\displaystyle {\binom {n-1}{k}}}

 as terms in the previous answers ? This is because the coefficient of

x

n
−
2

(
n
−
2
)
!

{\displaystyle {\frac {x^{n-2}}{(n-2)!}}}

 in the e.g.f. of F" is the same with coeff

x

n
−
1

(
n
−
1
)
!

{\displaystyle {\frac {x^{n-1}}{(n-1)!}}}

 for F' and with

x

n

n
!

{\displaystyle {\frac {x^{n}}{n!}}}

 for F even we do not have isomorphisms.

## 2990035

https://math.stackexchange.com/questions/2990035/probability-with-sets

The numbers from 1 to 10 are divvied randomly into 2 sets of 5 elements each.

1) In Species language, the structure in the problem is \*a 2-set of 5-sets of A or B or X

E

2

(

E

5

(
A
B
X
)
)

{\displaystyle E\_{2}(E\_{5}(A+B+X))}

 having the e.g.f

1

2
!

(

(
a
b
x

)

5

5
!

)

2

{\displaystyle {1 \over 2!}({\frac {(a+b+x)^{5}}{5!}})^{2}}

where X stands for the species of "1", Y for the species of "2" and Z for all other figures.

we are interested in the coefficient

a

1
!

b

1
!

x

8

8
!

{\displaystyle {a \over 1!}{b \over 1!}{x^{8} \over 8!}}

 which is 126.

There are then 126 way of partitioning 10 elements among 2 of them are marked.

The good configurations are

A
.
B
.

E

3

(
X
)
.

E

5

(
X
)

{\displaystyle A.B.E\_{3}(X).E\_{5}(X)}

The e.g.f is

a
.
b
.

x

3

3
!

x

8

8
!

{\displaystyle a.b.{x^{3} \over 3!}{x^{8} \over 8!}}

 and the coefficient of

a

1
!

b

1
!

x

8

8
!

{\displaystyle {a \over 1!}{b \over 1!}{x^{8} \over 8!}}

 is 56.

Then we have to apply the definition of a probability.