Sandrine has eight boxes numbered 1 to 8 and eight balls numbered 1 to 8. In how many ways can she put the balls in the boxes so that there is one ball in each box AND ball 1 is not in box 1 and ball 2 is not in box 2?
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There would be (assuming he places ball 1 first) a 7/8 chance of not putting the ball in box 1.
Likewise there would be a 6/7 chance not putting ball 2 in box 2.
Likewise for the remaining balls.
Therefore the probability of all balls not being placed in the box of the same number is: 7/8 x 6/7 x 5/6 x 4/5 x 3/4 x 2/3 x 1/2 = 1/8.
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