6p3
On Fri, Sep 7, 2012 at 7:36 PM, mohit mishra wrote:
> I think answer would be 6C3*3!. ie 120.
>
> On Fri, Sep 7, 2012 at 10:20 AM, Navin Kumar 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
I think answer would be 6C3*3!. ie 120.
On Fri, Sep 7, 2012 at 10:20 AM, Navin Kumar 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
>
@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 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 >wrote:
>
>>
120 arrangements...
-- RK :)
On Thu, Sep 6, 2012 at 6:00 PM, Navin Kumar wrote:
> @tendua: answer would be 6C3. Read about combination definition.
>
>
> On Thu, Sep 6, 2012 at 5:05 PM, atul anand wrote:
>
>> question says *3 alphabets with no data repeated* ...you no need of
>> doing 3! permut
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 wrote:
> seems output should be 20.
>
>
> On Thu, Sep 6, 2012 at 3:26 PM, tendua wrote:
>
>> from the set {a,b,c,d,e,f} find num
cmon.. what logic are you applying ? its the arrangements.. so basically
this is what will happen
there are three places __,__,__ first can be filled in 6 ways,Since
repetition is not allowed,second place could be filled in 5 ways and third
in 4
So total arrangements so far = 6*5*4 = 120 . Now t
Can you send the link to the question.
On Thu, Sep 6, 2012 at 4:35 PM, tendua 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 i
@tendua: answer would be 6C3. Read about combination definition.
On Thu, Sep 6, 2012 at 5:05 PM, atul anand 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 wrote:
>
>> fro
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 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 mu
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
seems output should be 20.
On Thu, Sep 6, 2012 at 3:26 PM, tendua 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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> "Algorit
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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