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permutation and combination

Asked by deepak.embedded 25th March 2010, 10:08 PM
Answered by Expert
Answer:

The total permutations of all five digits are 5! = 120.

So the arragements to be omitted are,

123**, 12**5, 1*3*5, 1**45,  **345, *234*, 1234*, *2345, 1*345, .....

And these corresponds to 2 arrangements each, therefore total 6x2 + 5 + 1 = 18

Hence the required number of arrangments = 120 - 18 = 102

Regards,

Team,

TopperLearning.

Answered by Expert 2nd April 2010, 7:02 AM
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