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Old 07-22-2008, 04:04 AM   #14 (permalink)
Goldust
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Join Date: May 2008
Location: India
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For arrangements with some items identical, there is a formula.

Suppose you have p items of which a items are identical, b other items are identical and so on till n identical items.

Then, the total number of arrangements is given by p!/(a!b!...n!)

In the first case (3-2-2), you have one 3 and two 2's. Hence, the total arrangements are 3!/2! = 3. Ditto for the second and fourth case where you have two 3's and two 1's respectively.

In the third case, you have 3 distinct elements and hence, the total arrangements are simply 3!
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