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A combinations with repetition

(a)Determine the number of integer solutions of

a1+a2+a3+a4=32

where a1,a2,a3>0, 0<a4<=25

(b)Mary has two dozen each of n different colored beads.

If she can select 20 beads(with repetition of colors allowed)

in 230,230 ways,what is the value of n?

Thanks

Update:

你好,第二題對了,但第一題ans=4475

Update 2:

and H 是何意,Thank you ^^

2 Answers

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  • 1 decade ago
    Favorite Answer

    (a) 求 a1 + a2 + a3 + a4 = 32 之所有未知數的正整數解。 題目改寫成 ( a1 - 1 ) + ( a2 - 1 ) + ( a3 - 1 ) + ( a4 - 1 ) = 28 求,所有 a1-1、a2-1、a3-1、a4-1 的非負整數解。即 H284 但 a4 ≦ 25 ⇒ a4 - 1 ≦ 24 。必須將多算得的組數減去。 當 a4 - 1 = 25 時,(a1-1) + (a2-1) + (a3-1) = 3 ,有 H33 組解。 當 a4 - 1 = 26 時,(a1-1) + (a2-1) + (a3-1) = 2 ,有 H23 組解。 當 a4 - 1 = 27 時,(a1-1) + (a2-1) + (a3-1) = 1 ,有 H13 組解。 當 a4 - 1 = 28 時,(a1-1) + (a2-1) + (a3-1) = 0 ,有 H03 組解。 所求 = H284 - H23 - H13 - H03 = C331 - C24 - C13 - C02    = 31.30.29 / 6 - 6 - 3 - 2 = 4484(b) Mary 有 n 種不同顏色的麵包,每種 2 打 24 個。現在拿 20 個麵包,  可重覆拿取,則共有 230,230 個拿法。問 n 的值。 按題意拿法有 H20n = C20n+20-1 = C20n+19 = Cn-1n+19  = (n+19) × ( n+18 ) × …… × 21 / ( n - 1 )! n = 1 ,所求 = 1 ; n = 2 ,所求 = 21 ; n = 3 ,所求 = 22×21 / 2! ; n = 4 ,所求 = 23×22×21 / 3! ; n = 5 ,所求 = 24×23×22×21 / 4! ; n = 6 ,所求 = P56+19 / 5! ; n = 7 時,所求 = P57+19 / 6! = 230,230

    2006-08-21 04:06:58 補充:

    Sorry for my inattention,第一題最後一行算錯了:H(4,28) - H(3,3) - H(3,2) - H(3,1) - H(3,0)= C(31,3) - C(5,3) - C(4,2) - C(3,1) - C(2,0)= 31.30.29 / 3! - 10 - 6 - 3 - 1 = 4495 - 10 - 6 - 3 - 1 = 4475

    2006-08-21 04:07:14 補充:

    H 是重覆組合(combination with repetition)的符號。在台灣,高中數學是用這個符號(至少我的高中是),外國的數學用何種符號我就不知道了。C(m,n) : 有 m 個東西,取 n 個之方法H(m,n) : 有 m 種東西,取 n 個且每種東西可重覆拿取之方法

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  • 1 decade ago

    這是重複組合的問題(H),算蠻基本的問題,

    建議你翻翻定義試試看喔。

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