There are several sets of M,N with resolution for the set M,N satisfying M N a,b,c

1) m= There is only one n=

2) m=n must have at least b and c, then there are two kinds of sums.

3) In the same way, there is m=

n must have at least a and c, then there are two kinds of sums.

4) In the same way, there is m=

n must have at least a and b, then there are two sums.

5) m=n must have at least c, then there is and, 6) m=

n must have at least b, then there is and, 6) m=

n must have at least a, then there is and, 7) m=

Then n does not exceed the can, ,

Solution: n=. when m=

When m=n=or.

n= when m=, n= or, when m=.

When m=n=or.

When m=n=or.

When m=n=or.

When m=n=or.

In total, there are 29 types.

a}.Considering the substarved fiber order, there are 2 groups).

a,b} consider the order of prudence, then there are 2 groups) a,b} consider the order, there are 2 groups).

a,b} If only combinations are considered, there are 5 groups.

If the order m and n are considered separately, there are 9 groups.

m=,n=,dust accompanies m n=,m n=,m n,m n,m n, then the correct formula for the skin of the ant is b, so choose b

Obviously, m= is contained in, so as long as the non-empty set n is also contained in, mun must be contained in.

That is, the non-empty set n is a true subset, so the number of n is 2 3 -1 = 7 collapses.

n can be equal to 、、、 sedan posture, ,

Solution: The set m=,n= has three elements.

There are 2 subsets of m.

ps: If a set contains n elements, then the number of its subsets is 2, and the n proof is calculated according to the number of elements.

A subset does not contain elements, and there is an empty set.

The subset contains an element of c(1,n) species.

There are two elements in the subset and there are c(2,n) species.

There are n elements in the subset, there are c(n,n) species, and the number of subsets can be obtained as c(0,n)+c(1,n)+....c(n,n)=2^n.

Solve 0 elements - 1 empty set.

c31 = 3 for 1 element

c32 = 3 for 2 elements

C33 = 1 for 3 elements

Therefore, the number of elements in the set n is at most 1+3+3+1=8, which seems to be difficult to summarize directly to the 3rd power of 2.

Because the question asks how many are at most, and there are a total of 8 empty sets you listed, and 8 is the third power of 2, which is correct.

I'll give you an example: a= b= has an empty set.

That is, to the nth power of 2, n represents the number of elements.

m∪n=;P Kona can be empty confession set, a hole to lift the base is a total of 8 possibilities.

The set m,n that satisfies m n= has: m= ,n=; m=，n=；Nian Xiao m=,n=; m=，n=；m=，n=； m=，n=∅；m=，n=；m=，n=；m=,n= There are 9 groups so choose c