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# High School Mathematics Competition (00-06) Examination Questions Summary-Permutations and Combinations

High School Mathematics Competition (00-06) Examination Questions Summary-Permutations and Combinations

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Topic VII Permutations, Combinations, Binomial Theorems and Probabilities

One,

Multiple-choice questions (6 points each)

21000

2

2000

1. (01 country) If the expansion of (1 + x + x) is a0 + a1x + a2x ++ a2000x, then a0 + a3 + a6 +

The value of a9 ＋＋ a1998 is () A.3 B3 C.3 D.3

333

666

999

2001

Solution: Since the sum of two coefficients of each drop in the expansion is required, the unit root of 1 is associated with the special value method.

Take ω =-(1/2) + () Let x = 1, get 3

1000

/ 2) i, then ω = 1, ω + ω + 1 = 0. = A0 + a1 + a2 + a3 + + a2000; ①

2

2000

32

Let x = ω, and get 0 = a0 + a1ω + a2ω + + a2000ω

2

2

4

6

; ②

4000

Let x = ω, get 0 = a0 + a1ω + a2ω + a3ω + + a2000ω  + + + + ③ get 3

1000

． ③

= 3 (a0 + a3 + a6 + + a1998).

999

∴a0 + a3 + a6 + + a1998 = 3, choose C.

2. (Nationwide 02) Given two sets of real numbers A = ｛a1, a2,, a100 ｛and B = ｛b1, b2,, b50｝, if the mapping f from A to B makes every element in B have the original , And f (a1) ≤f (a2) ≤ ≤f (a100), such mappings share ()

50100

4899

49

49

(A) C (B) C (C) C

100

(D) C

99

Solution: Let b1 <b2 << b50 and divide the elements a1, a2, and a100 in A into 50 non-empty groups in order.

Define the mapping f: A → B, so that the elements of the i-th group under f are all bi (i = 1,2,, 50). It is easy to know that such f meets the requirements of the problem, and each such group is One-to-one mappings that satisfy the conditions, so the number of mappings f that meet the design requirements is equal to the number of divisions in which A is divided into 50 groups in full code order, and the number of divisions in A is C, so D is selected.

4999

, Then such a mapping has a total of C

4999

,

3． (Nationwide 04) Set three digits n abc. If a, b, and c are the length of three sides to form an isosceles (including isosceles) triangle, then such three digits n have () A. 45

B. 81

C. 165

D. 216

Solution: if a, b, and c can form the side length of the triangle, obviously they are not zero. That is, a, b, c {1,2, ..., 9}. (1) If an equilateral triangle is formed, let the number of such three digits be n1. Since the three digits in the three digits are all the same, so