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? >>>>>> >>>>>> -- >>>>>> You received this message because you are subscribed to the Google >>>>>> Groups "Algorithm Geeks" group. >>>>>> To view this discussion on the web visit https://groups.google.com/d/ >>>>>> **ms**g/algogeeks/-/E4U2XlfkvgMJ<https://groups.google.com/d/msg/algogeeks/-/E4U2XlfkvgMJ> >>>>>> . >>>>>> To post to this group, send email to [email protected]. >>>>>> To unsubscribe from this group, send email to algogeeks+...@** >>>>>> googlegroups.com**. >>>>>> >>>>>> For more options, visit this group at http://groups.google.com/** >>>>>> group**/algogeeks?hl=en<http://groups.google.com/group/algogeeks?hl=en> >>>>>> . >>>>>> >>>>> >>>>> -- >>>> You received this message because you are subscribed to the Google >>>> Groups "Algorithm Geeks" group. >>>> To view this discussion on the web visit https://groups.google.com/d/** >>>> msg/algogeeks/-/VMO1othQRcQJ<https://groups.google.com/d/msg/algogeeks/-/VMO1othQRcQJ> >>>> . >>>> >>>> To post to this group, send email to [email protected]. >>>> To unsubscribe from this group, send email to algogeeks+...@** >>>> googlegroups.com. >>>> For more options, visit this group at http://groups.google.com/** >>>> group/algogeeks?hl=en <http://groups.google.com/group/algogeeks?hl=en>. >>>> >>> >>> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To view this discussion on the web visit >> https://groups.google.com/d/msg/algogeeks/-/z6KbH2i6BP0J. >> >> To post to this group, send email to [email protected]. >> To unsubscribe from this group, send email to >> [email protected]. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
