# F.5 math~probability(2)

Six balls A, B, C, D, E and F are put into six boxes numbered 1, 2, 3, 4, 5, and 6. If only one ball can be put into each box, what is the probability that ball A is not in box 1 and ball B is not in box 6?

http://hk.knowledge.yahoo.com/question/question?qi...

這條題目之前有人答過..但我唔明佢既計法

[4*4+1*5]*4*3*2*1係點樣諗出黎?

或者有冇易明D既方法?

### 1 Answer

Rating

- myisland8132Lv 79 years agoFavorite Answer
Just using the inclusion-exclusion principle is OK. The no. of choice

= N(no restriction) - N(A is in box 1) - N(B is in box 6) + N(A and B is in box 6)

= 6! - 5! - 5! + 4!

= 504

So, P(ball A is not in box 1 and ball B is not in box 6)

= 504/720

= 7/10

Hope that helps !

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