http://placement.freshersworld.com/placement-papers/Persistent-/Placement-Paper-Whole-Testpaper-1884
question no. 4 in 5th section

On Thursday, September 6, 2012 4:40:08 PM UTC+5:30, isandeep wrote:
>
> Can you send the link to the question.
>
> On Thu, Sep 6, 2012 at 4:35 PM, tendua <[email protected] 
> <javascript:>>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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