Combinations with Repetition?

2015-12-11 9:24 am
x1+x2+x3=7 where x1,x2,x3 >= 0
How many positive integer solutions are there?
更新1:

thank you!! I understand the method 1, but i have no idea with method 2. can you explain more?

回答 (1)

2015-12-11 10:34 am
更新:
題目問 「How many "positive integer solutions" are there?」
即 x₁, x₂, x₃ ≠ 0 嘅解法 ?

如果係嘅話
Method 1
x₁ + x₂ + x₃ = 7, x₁, x₂, x₃ ≠ 0
_, _, _, _, _, _, _, |, | 嘅排法 ( 但 |, | 唔可以排埋一齊 )
∴ C(6,2) = 15

Method 2
y₁ + y₂ + y₃ = 4, y₁, y₂, y₃ ≥ 0
[ x₁ = y₁ + 1, x₂ = y₂ + 1, x₃ = y₃ + 1 ]
H(3,4) = C(4+3-1,4) = C(4+3-1,3-1) = 15

∴ 15


若 x₁, x₂, x₃ 可以 = 0
x₁ + x₂ + x₃ = 7
Method 1:
1.
_ _ | | _ _ _ _ _:x₁ = 2, x₂ = 0, x₃ = 5
2.
| _ _ _ _ | _ _ _:x₁ = 0, x₂ = 4, x₃ = 3
......
∴ We need to permute _, _, _, _, _, _, _, |, |
But _, _, _, _, _, _, _ are the same
Also, |, | are the same

∴ P(9,2) P(7,7) / (2!7!) = P(9,7) P(2,2) / (2!7!) = 36
[ C(9,2) = C(9,7) = 36 ]


Method 2:
H(3,7) = C(7+3-1,7) = C(7+3-1,3-1) = 36

∴ 36


收錄日期: 2021-04-18 14:15:10
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151211012440AAFW0xq

檢視 Wayback Machine 備份