permutation and combination
Asked by deepak.embedded | 25th Mar, 2010, 10:08: PM
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 | 2nd Apr, 2010, 07:02: AM
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