probability

Asked by  | 12th Feb, 2010, 08:12: PM

Expert Answer:

Let M, and P denote the books of mathematics and physics respectively.

Now, all the mathematics books can be placed side by side, in only following configurations.

M1M2M3M4M5P1P2P3

P1M1M2M3M4M5P2P3

P1P2M1M2M3M4M5P3

P1P2P3M1M2M3M4M5

Total 8 books can be permuted or arranged on the bookself in 8! ways.

Let's count the total permutations in which the mathematics books are side by side.

For 1st arrangemnet indicated above, it's 5!3!

For 2nd arrangement, it's 5!2!, but it's P1 in first position and we can have P2 and P3 as well.

Therefore for 2nd arrangement, it's 3x5!2!.

For 3rd arrangement, same ways as done for 2nd, only we have one book at the end now,

Therefore for 3rd also, it's 3x5!2!.

For 4th it's same as the 1st case, so 5!3!

The probability that mathematics books are placed side by side is = (2x5!3! + 6x5!2!)/8!

= (2x5!2!x3 + 6x5!2!)/8! = 12(5!2!)/8! = 12x2/(6x7x8) = 1/14

Regards,

Team,

TopperLearning.

 

Answered by  | 13th Feb, 2010, 07:11: AM

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