AMC 8 2004 Problem 17
AMC8 2004 年第 17 题
排列与组合★★★☆☆
Three friends have a total of 6 identical pencils, and each one has at least one pencil. In how many ways can this happen?
3 个朋友总共有 6 支一模一样的铅笔，并且每个人都有至少一支铅笔。那么一共有多少种可能的分配铅笔的方法？
- A.1
- B.3
- C.6
- D.10
- E.12
正确答案 · Correct Answer
D
解析 · Solution
Solution 1
For each person to have at least one pencil, assign one pencil to each of the three friends so that you have 3 left. In partitioning the remaining 3 pencils into 3 distinct groups, use Ball-and-urn to find the number of possibilities is . Solution by phoenixfire Minor Edit by Yuvag
Like in solution 1, for each person to have at least one pencil, assign one of the pencils to each of the three friends so that you have 3 left. In partitioning the remaining 3 pencils into 3 distinct groups, use number of non-negative integral solutions. Let the three friends be a, b, c respectively. a+b+c = 3 The total being 3 and 2 plus signs, which implies . Solution by phoenixfire Minor Edit by Yuvag
For each of the 3 People to have at least one pencil when distributing 6 pencil amongst them, we can use another formula from the Ball-and-urn counting technique, shown below: for n = number of items, and s = slots: Now we can plug in our values, number of items = 6, and slots = 3: . Solution by Yuvag
Like in solution 1 and solution 2, assign one pencil to each of the three friends so that you have 3 left. In partitioning the remaining 3 pencils into 3 distinct groups use casework. Let the three friends be , , repectively. a+b +c = 3, Case 1:a=0, b+c = 3, b = 0,1,2,3, c = 3,2,1,0, solutions. Case 2:a=1, 1 + b + c = 3, b + c = 2, b = 0,1,2, c = 2,1,0, solutions. Case 3:a= 2, 2 + b + c = 3, b + c = 1, b = 0,1, c = 1,0, solutions. Case 4:a = 3, 3+b+c = 3, b+c = 0, b = 0, c = 0, solution. Therefore there will be a total of 4+3+2+1=10 solutions. Solution by phoenixfire
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