Questions about permutation 急!

Leung

2009-07-23 9:32 pm

1. How many different six-digit numbers can be formed using digits 2,3,3,3,4,4 ? How many of there are even?

2.In how many ways can six children form a circle to play a game? If Joan refuses to stand next to Tom, how many possible arrangements are the?

更新1:

myisland Can you please answer me one more question??? Find the number of permutations of the word PARABOLA. In how many of there permutations are (a) all three As together? (b) no two As together? If you answer this questions for me. I will give you the 20marks immediately!

回答 (1)

myisland8132

2009-07-23 11:31 pm

✔ 最佳答案

1 There are 1 2's, 3 3's and 2 4's. Using formula, the combinations is
6!/1!3!2!=720/12=60
2 There are (6-1)!=120 ways for six children to form a circle to play a game.
If we treat Joan and Tom to be the one. Then there are 4!=24 way to form a circle. Of course there are 2!=2 ways for Joan and Tom to change their places. So, the no. of ways is 120-2*24=120-48=72

2009-07-23 17:24:05 補充：
PARABOLA, 3 A's, 1 B's, 1 L's, 1O's, 1 P's, 1 R's

(a) Treat three A's as one and using the formula

6!/1!1!1!1!1!1!=720

2009-07-23 17:24:27 補充：
(b) Let say A1A2 are combined together, called it S, then the number of ways to make a permutations of the word PARABOLA is such that S and A3 are not together is

7!-2*6!=3600

2009-07-23 17:24:30 補充：
Similarly for A2A3 or A1A3 combined together

Using the result of (a) and inclusion -exclusion principle

The possible no of ways is

(8!-2*3*3600)/6
=3120

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