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combinatorics 2

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by surajgarg » Wed Jul 28, 2010 12:38 pm
The middle 3 digits can be any digit from 0 to 9.

That means - 10x10x10 = 1000

The first and last digits can have combination as

2 _ _ _ 2
2 _ _ _ 4
4 _ _ _ 4
4 _ _ _ 2

4 combinations.

Total 5 digit combinations possible = 1000 x 4 = 4000
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by nicolezl » Wed Jul 28, 2010 12:40 pm
There are 2 options for the first and last digit, and 10 options for the middle 3, so total combinations=

2 X 10 X 10 X 10 X 2 = 4000
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by Gurpinder » Wed Jul 28, 2010 12:40 pm
ahh thanks. i was looking at it from a completely different perspective.
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