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permutation and combination

Asked by deepak.embedded 25th March 2010, 10:09 PM
Answered by Expert
Answer:

possible triplets (x,y,z) such that xyz=288  since you have not specified about x,y,z to be distinct so we are considering al the cases

When 2 numbers are distinct then the ordered pair can be arranged in 3 ways and in case of all 3 distinct they can be aranged in 6 ways.

(2,12,12), (4,4,18), (9,2,16), (6,6,8),(3,3,32),(3,8,12),(4,3,24),(2,8,18)

 

Answered by Expert 30th March 2010, 8:00 PM
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