6p3

On Fri, Sep 7, 2012 at 7:36 PM, mohit mishra <[email protected]> wrote:

> I think answer would be 6C3*3!. ie 120.
>
> On Fri, Sep 7, 2012 at 10:20 AM, Navin Kumar <[email protected]>wrote:
>
>> @tendua: Answer will be 6C3 x 3! .
>>
>> For example: If 5 letters are given then you can get only 10 combination
>> of different letter = 5C3
>>
>> ABC
>> ABD
>> ABE
>> BCD
>> BCE
>> CDE
>> ACD
>> ACE
>> ADE
>> BDE
>>
>> now each of these can be arranged in 3! ways. So final answer will be :
>> 120
>>
>>
>> On Fri, Sep 7, 2012 at 1:11 AM, tendua <[email protected]> wrote:
>>
>>>
>>> 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]> 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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-- 
*Anil  Arya,
Computer Science *
*Motilal Nehru National Institute of Technology,Allahabad .
*

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