緊急!有關Permutation&Combination

[匿名]

2010-12-15 5:47 am

A standard pack of playing cards consists of 4 suits (Clubs, Diamonds, Hearts, Spades), each of 13 cards numbered in ascending order as 2,3,4, ..., 10, Jack, Queen, King, Ace. 5 cards are dealt from the standard pack. In how many ways can the following be drawn:
a. A full-house
b. exactly one pair
c. exactly two pairspart a其實我唔係好明咩野係Full house
答案就寫13C1x4C3x12C1x4C2
乜唔係應該13C2x4C3x4C2架咩?
我知道計出黎既答案唔同..但依兩條式有咩分別?part b 答案寫13C1x4C2x[(48C1x44C1x40C1)]/3!
依度點解要除3!架??part c同part a一樣..但個答案反而係13C2x4C2x4C2x44C1
而唔係13C1x4C2x12C1x4C2x44C1
點解同part a唔同既?

回答 (1)

用戶9112

2010-12-16 5:41 am

✔ 最佳答案

(a) Full house(夫佬)即三隻一樣加二隻一樣.13C2 = 13個數字中選兩個(如A+K),但AAAKK及KKKAA是不一樣的.做法應該是先選兩個數字,計算A是3隻的情況,然後計算K是3隻的情況.所以number of ways = (13C2)[(4C3)(4C2)+(4C3)(4C2)](b) (48C1)(44C1)(40C1)是說選擇3隻不同數字的情況,如AKJ但選擇順序可以是:AKJ,AJK,KAJ,KJA,JAK或JKA有3!=6次重複,所以要除以6另外也可以這樣考慮,先選3個字,再乘以毎個數字的可能性,從而得出(13C1)(4C2)(12C3)(4C1)(4C1)(4C1)(c) 這一次的情況跟夫佬不同,AAAKK不等於AAKKK,但AAKK=KKAA若你計算用(13C1)(4C2)(12C1)(4C2)(44C1)的話,就會出現AAKK,KKAA的重複情況.

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