1, Question: If at least one letter is extracted from letters A, B and C, how many possible extraction results are there? List possible outcomes.

Solution: The classification is discussed as follows:

1 > When extracting a letter, * * has 3C 1=3 possible extraction results, namely A, B and C.

2 > When extracting two letters, * * * has 3C2=3 possible extraction results, as shown below:

AB，BC，CA

3 > extract three letters, * * * has 3C3= 1 extraction results, namely ABC.

Note: the combination number formula ncm = [n (n-1) * ... (n-m+1)]/m!

2. Question: Let the set S={A, b, C}

Question 1: If at least one element is extracted from set S to form set T, what is the number of sets T that meet this condition? And list them one by one.

Question 2: Let the elements of set M be all sets T in 1. Now, several elements are extracted from the set M to form the set N. If the set N satisfies that the union of its elements is S and the intersection of its elements is not equal to S, find the number of the set N and list them one by one.

Question 1: Same as the previous question.

Solution to the second problem: According to the solution and meaning of the first problem, the set T = {{a}, {b}, {c}, {ab}, {AC}, {BC} and {ABC} are classified and discussed as follows:

1 > when the set n contains an element, obviously only N={{ABC}} satisfies the problem.

2 > when the set n contains two elements, if one element itself contains one element, then the other element can only contain two elements, otherwise the necessary condition that the intersection of elements in n is not equal to S is not satisfied, and the possible results are {{a}, {b, c}}, {{c}, {

If two elements of n contain at least two elements, then the above two elements can only contain two elements, namely {{a, b}, {b, c}}, {{a, b}, {a, c}}, {{a, c}, {b, c}}.

3 > When the set n contains three elements, obviously the results can only be {{A}, {B}, {C}} and {{A, B}, {B, C}, {A, C}}.

To sum up, a * * * has nine sets n that meet the topic, namely: {{ABC}},

{{A}、{B、C}}、{{B}、{A、C}}、{{C}、{A、B}}、{B、C}}、{{A、B}、{A、C}}、{{A、C}、{B、C}}、{{A}、{B}、{C}}、{{A }、{B}、{ C} }、{ {A、B}、{ B }、{ B }、{ A、C }。

Note: this problem can be solved by permutation and combination formula in one step, but because the questioner needs a detailed solution process, I use the method of classification and addition. Although I don't blatantly use the so-called permutation and combination formula, my thinking is clear and easy to understand. In fact, the formula is derived from the above process.