120 arrangements...

--  RK :)


On Thu, Sep 6, 2012 at 6:00 PM, Navin Kumar <[email protected]>wrote:

> @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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