✔ 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
