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
  • 9 years ago
    Favorite 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 !

Still have questions? Get your answers by asking now.