@tendua: answer would be 6C3. Read about combination definition.

On Thu, Sep 6, 2012 at 5:05 PM, atul anand <[email protected]> wrote:

> question says *3 alphabets with no data repeated* ...you no need of doing
> 3! permutation.
> eg 123 and 321 are same
>
>
> On Thu, Sep 6, 2012 at 4:35 PM, tendua <[email protected]> wrote:
>
>> from the six elements, we could choose any three in C(6,3) ways which is
>> 20 and then permute all the three elements so it will be multiplied by 3!
>> which is 6. Hence, 20*6 = 120. We still have to multiply it by 3 to get 360
>> but I'm not getting why?
>>
>>
>> On Thursday, September 6, 2012 3:54:11 PM UTC+5:30, atul007 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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