I guess 720 must be the ans.
Its like forming three-letter words out of given 6 without repetition.
In that case = 6p3 * 3!

On Thu, Sep 6, 2012 at 3:53 PM, atul anand <[email protected]> wrote:

> seems output should be 20.
>
>
> On Thu, Sep 6, 2012 at 3:26 PM, tendua <[email protected]> wrote:
>
>> from the set {a,b,c,d,e,f} find number of arrangements for 3 alphabets
>> with no data repeated?
>> Answer given is 360. but how?
>>
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-- 
best wishes!!
 Vaibhav

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