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標題:
nCr,nPr
發問:
1. Nancy has bought 10 different christmas presents. In how many ways can she (a) divide the present into 5 groups of 2? (b) give all the presents to 2 friends such that each of her friends receives at least 3 presents?2. A 5-card poker hand is selected from a deck of 52 playing cards. How many different poker... 顯示更多 1. Nancy has bought 10 different christmas presents. In how many ways can she (a) divide the present into 5 groups of 2? (b) give all the presents to 2 friends such that each of her friends receives at least 3 presents? 2. A 5-card poker hand is selected from a deck of 52 playing cards. How many different poker hands contain exactly one pair? Thank you indeed.
最佳解答:
1a)10! / (2! * 2! * 2! * 2! * 2! * 5!) = 945 ways or (10C2)(8C2)(6C2)(4C2)(2C2) / 5! = 945 waysb)Divide the present into 2 groups of 3&7 or 7&3 = (10C3)(7C7) + (10C7)(7C3) = 120 + 120 = 240 ways 2 groups of 4&6 or 6&4 = (10C4)(6C6) + (10C6)(4C4) = 210 + 210 = 420 ways 2 groups of 5&5 = (10C5)(5C5) = 252 ways The required number of ways = 240 + 420 + 252 = 912 ways. 2)(52C1 * 3C1) * (48C1 * 44C1 * 40C1) / (2! 3!) = 1098240 waysor 13C1 * 4C2 * 12C3 * 43 = 1098240 ways 2012-03-02 17:57:35 補充: Typing error : (10C3)(7C7) + (10C7)(7C3) = 120 + 120 = 240 ways should be (10C3)(7C7) + (10C7)(3C3) = 120 + 120 = 240 ways
其他解答:
nCr,nPr
發問:
1. Nancy has bought 10 different christmas presents. In how many ways can she (a) divide the present into 5 groups of 2? (b) give all the presents to 2 friends such that each of her friends receives at least 3 presents?2. A 5-card poker hand is selected from a deck of 52 playing cards. How many different poker... 顯示更多 1. Nancy has bought 10 different christmas presents. In how many ways can she (a) divide the present into 5 groups of 2? (b) give all the presents to 2 friends such that each of her friends receives at least 3 presents? 2. A 5-card poker hand is selected from a deck of 52 playing cards. How many different poker hands contain exactly one pair? Thank you indeed.
最佳解答:
1a)10! / (2! * 2! * 2! * 2! * 2! * 5!) = 945 ways or (10C2)(8C2)(6C2)(4C2)(2C2) / 5! = 945 waysb)Divide the present into 2 groups of 3&7 or 7&3 = (10C3)(7C7) + (10C7)(7C3) = 120 + 120 = 240 ways 2 groups of 4&6 or 6&4 = (10C4)(6C6) + (10C6)(4C4) = 210 + 210 = 420 ways 2 groups of 5&5 = (10C5)(5C5) = 252 ways The required number of ways = 240 + 420 + 252 = 912 ways. 2)(52C1 * 3C1) * (48C1 * 44C1 * 40C1) / (2! 3!) = 1098240 waysor 13C1 * 4C2 * 12C3 * 43 = 1098240 ways 2012-03-02 17:57:35 補充: Typing error : (10C3)(7C7) + (10C7)(7C3) = 120 + 120 = 240 ways should be (10C3)(7C7) + (10C7)(3C3) = 120 + 120 = 240 ways
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