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Re: [Ascent CAT] try to solve-permutation and combination-imp

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  • Nirav khandhadia
    Hi mandar, the answer to the second question: nC2 = 45 = n!/(2!)(n-2)! = n(n-1) = 90 = n(n-1) = 10X 9 so, n=no. of teams = 10. bye; mandar keny
    Message 1 of 13 , Mar 3, 2006
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      Hi mandar,
      the answer to the second question:
      nC2 = 45 => n!/(2!)(n-2)! = n(n-1) = 90 => n(n-1) = 10X 9
      so, n=no. of teams = 10.
      bye;


      mandar keny <mandar_gec@...> wrote:
      1. in a number consisting of 7 digits i.e 4 zeros and
      3 threes. we need to form a 7 digit no. how?
      pllzz explain if anyone can... ans is 15

      2.in a competition a several teams take part. each
      team plays against all opp team. total no games played
      is 45. how many teams are there?
      ans given is 10
      bt hw?



                 
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    • anil kuppa
      Hi mandar the answer to the first question: The first place can be occupied by only 3 since it cannot be 0. 6 places are left now.Here you need to do the
      Message 2 of 13 , Mar 5, 2006
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        Hi mandar
         the answer to the first question:
          The first place can be occupied by only 3 since it cannot be 0.
        6 places are left now.Here you need to do the selection.There are 6 places out of which there are 4 zeroes and 2 threes left.You can do it in 6c2 ways=15.
        Hope you understood it.Whenever there is a repetition you should do combination since the order is not important and all of them have to be in the number anways.
           OR
           First place can be filled in one way.
        There are 6 places left with 4 0's and 2 3's.
        There is a formula for that
         n!/(p!.q!.....)
         where 'n' are the places .
         P and Q are the  repetitions.


         
        On 3/3/06, Nirav khandhadia <nirav_khandhadia2004@...> wrote:

        Hi mandar,
        the answer to the second question:
        nC2 = 45 => n!/(2!)(n-2)! = n(n-1) = 90 => n(n-1) = 10X 9
        so, n=no. of teams = 10.
        bye;


        mandar keny <mandar_gec@...> wrote:
        1. in a number consisting of 7 digits i.e 4 zeros and
        3 threes. we need to form a 7 digit no. how?
        pllzz explain if anyone can... ans is 15

        2.in a competition a several teams take part. each
        team plays against all opp team. total no games played
        is 45. how many teams are there?
        ans given is 10
        bt hw?



                   
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      • Abhijeet Singh
        Hi mandar, the ans to ur 1st ques is 1st digit of the 7-digit number cannot be 0, therefore it is 3 for the rest of the 6 positions we have 4 0 s and 2
        Message 3 of 13 , Mar 6, 2006
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          Hi mandar,
          the ans to ur 1st ques is
          1st digit of the 7-digit number cannot be 0, therefore it is 3
          for the rest of the 6 positions we have 4   0's and  2   3's.
          therefore, number of combinations is 6!/(4!x2!) = 15


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        • GNANESHWAR KAKDE
          Q1 first place can be filled in 1 way ( Only 3 can occupy the plaace) balance 6 places can befilled by 6! ways 0 is repeated 4 times and balance 3 s is
          Message 4 of 13 , Mar 6, 2006
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            Q1
             
            first place can be filled in 1 way ( Only 3 can occupy the plaace)
             
            balance 6 places can befilled by 6! ways
             
            0 is repeated 4 times and balance 3's is reapeted 2 times so
             
            6!/(4!*2!) will be the final answer
             
            Regards

            Nirav khandhadia <nirav_khandhadia2004@...> wrote:

            Hi mandar,
            the answer to the second question:
            nC2 = 45 => n!/(2!)(n-2)! = n(n-1) = 90 => n(n-1) = 10X 9
            so, n=no. of teams = 10.
            bye;


            mandar keny <mandar_gec@...> wrote:
            1. in a number consisting of 7 digits i.e 4 zeros and
            3 threes. we need to form a 7 digit no. how?
            pllzz explain if anyone can... ans is 15

            2.in a competition a several teams take part. each
            team plays against all opp team. total no games played
            is 45. how many teams are there?
            ans given is 10
            bt hw?



                       
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          • rimpu varshney
            Hi Mandar, For QUESTION 2 : The formula comes upto : (n*2 - n) / 2 = no of games played. = (n*2 - n) / 2 = 45 = (n - 10) (n + 9) = 45 = n = 10 Regards,
            Message 5 of 13 , Mar 9, 2006
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              Hi Mandar,
               
              For QUESTION 2 :
               
              The formula comes upto  : (n*2  - n) / 2 = no of games played.
               
              => (n*2  - n) / 2 = 45
              => (n - 10) (n + 9) = 45
              => n = 10
               
              Regards,
              Rimpu Varshney

              mandar keny <mandar_gec@...> wrote:
              1. in a number consisting of 7 digits i.e 4 zeros and
              3 threes. we need to form a 7 digit no. how?
              pllzz explain if anyone can... ans is 15

              2.in a competition a several teams take part. each
              team plays against all opp team. total no games played
              is 45. how many teams are there?
              ans given is 10
              bt hw?



                         
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            • phanindra H
              Hi mandar here is the answer for your question: to form a seven digit number first place should be filled by 3 this can be done in only one way as there are
              Message 6 of 13 , Mar 11, 2006
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                Hi mandar
                here is the answer for your question:
                 
                                          to form a seven digit number first place should be filled by 3 this can be done in only one way as there are only threes and zeros and the ramaining six places can be filled in 6!/(2!*4!) ways that is the answer is 1*[6!/(2!*4!)] ie 15 ways we have divided 6! by (2!*4!) because as there are 2 threes and 4 zeros left and selecting any three among the two threes or selecting any zero among four zeros is immaterial
                rimpu varshney <rimpuv_2000@...> wrote:
                Hi Mandar,
                 
                For QUESTION 2 :
                 
                The formula comes upto  : (n*2  - n) / 2 = no of games played.
                 
                => (n*2  - n) / 2 = 45
                => (n - 10) (n + 9) = 45
                => n = 10
                 
                Regards,
                Rimpu Varshney

                mandar keny <mandar_gec@...> wrote:
                1. in a number consisting of 7 digits i.e 4 zeros and
                3 threes. we need to form a 7 digit no. how?
                pllzz explain if anyone can... ans is 15

                2.in a competition a several teams take part. each
                team plays against all opp team. total no games played
                is 45. how many teams are there?
                ans given is 10
                bt hw?



                           
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              • krippashankar
                Hi, Answer for the first question is: since we have 4 zero s and 3 three s we have two distinct numbers(0 and 3) to make a 7 digit number from them we will
                Message 7 of 13 , Mar 13, 2006
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                  Hi,
                  Answer for the first question is:
                  since we have 4 zero's and 3 three's
                  we have two distinct numbers(0 and 3)
                  to make a 7 digit number from them we will require to have
                  3 to take the first place which can be done in 1way and the
                  remaining paces will taken by the remaining numbers in 6! ways.
                  but we have 4 0's and 2 three's to occupy 6 places.hence the total
                  number of ways are (1*6!)/(2!*4!)=15





                  --- In ascent4cat@yahoogroups.com, rimpu varshney <rimpuv_2000@...>
                  wrote:
                  >
                  > Hi Mandar,
                  >
                  > For QUESTION 2 :
                  >
                  > The formula comes upto : (n*2 - n) / 2 = no of games played.
                  >
                  > => (n*2 - n) / 2 = 45
                  > => (n - 10) (n + 9) = 45
                  > => n = 10
                  >
                  > Regards,
                  > Rimpu Varshney
                  >
                  > mandar keny <mandar_gec@...> wrote:
                  > 1. in a number consisting of 7 digits i.e 4 zeros and
                  > 3 threes. we need to form a 7 digit no. how?
                  > pllzz explain if anyone can... ans is 15
                  >
                  > 2.in a competition a several teams take part. each
                  > team plays against all opp team. total no games played
                  > is 45. how many teams are there?
                  > ans given is 10
                  > bt hw?
                  >
                  >
                  >
                  >
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                • love_vineet13
                  mandar has given correct solution and the second question solution is as foolows n combination 2 = 45 ie n!/(n-2)!2!=45 n(n-1)=90 this gives n = 10 ... place
                  Message 8 of 13 , Mar 18, 2006
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                    mandar has given correct solution and the second question solution
                    is as foolows
                    n combination 2 = 45
                    ie n!/(n-2)!2!=45
                    n(n-1)=90
                    this gives n = 10


                    --- In ascent4cat@yahoogroups.com, phanindra H <phani_acumen@...>
                    wrote:
                    >
                    > Hi mandar
                    > here is the answer for your question:
                    >
                    > to form a seven digit number first
                    place should be filled by 3 this can be done in only one way as
                    there are only threes and zeros and the ramaining six places can be
                    filled in 6!/(2!*4!) ways that is the answer is 1*[6!/(2!*4!)] ie 15
                    ways we have divided 6! by (2!*4!) because as there are 2 threes and
                    4 zeros left and selecting any three among the two threes or
                    selecting any zero among four zeros is immaterial
                    > rimpu varshney <rimpuv_2000@...> wrote:
                    > Hi Mandar,
                    >
                    > For QUESTION 2 :
                    >
                    > The formula comes upto : (n*2 - n) / 2 = no of games played.
                    >
                    > => (n*2 - n) / 2 = 45
                    > => (n - 10) (n + 9) = 45
                    > => n = 10
                    >
                    > Regards,
                    > Rimpu Varshney
                    >
                    > mandar keny <mandar_gec@...> wrote:
                    > 1. in a number consisting of 7 digits i.e 4 zeros and
                    > 3 threes. we need to form a 7 digit no. how?
                    > pllzz explain if anyone can... ans is 15
                    >
                    > 2.in a competition a several teams take part. each
                    > team plays against all opp team. total no games played
                    > is 45. how many teams are there?
                    > ans given is 10
                    > bt hw?
                    >
                    >
                    >
                    >
                    > __________________________________________________________
                    > Yahoo! India Matrimony: Find your partner now. Go to
                    http://yahoo.shaadi.com
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                  • suvarchalarani
                    2.in a competition a several teams take part. each ... Hi Mandar If u got the explanation to this as to how you arrive at nC2 then, fine. Otherwise, think on
                    Message 9 of 13 , Mar 22, 2006
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                      2.in a competition a several teams take part. each
                      > team plays against all opp team. total no games played
                      > is 45. how many teams are there?
                      > ans given is 10
                      > bt hw
                      ?


                      Hi Mandar

                      If u got the explanation to this as to how you arrive at nC2 then, fine.

                      Otherwise, think on these lines

                      A game could be played between any 2 teams, right?
                      when there are n teams , everytime 2 teams are selected in nC2 ways

                      given that 45 games are played
                      so, nC2=45

                      If you know by experience, that nC2=45 when n=10, good.

                      Otherwise, nC2=n(n-1)(n-2)!/(n-2)!2!

                      =>n(n-1)=45*2=90
                      =>n(n-1)=10*9
                      =>n=10

                      Or by lenghty method, solve the quadratic equation that results from
                      the above as:
                      n^2-n-90=0

                      Discard the negative value for n and consider the positive value only

                      Hope this explanation helped you

                      Good luck

                    • vineet kumar
                      if u don t wanna apply any formula then u can do it by basics......but it takes time.... solution: suppose there are n teams playing in the competetion name
                      Message 10 of 13 , Mar 30, 2006
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                        if u don't wanna apply any formula then u can do it by basics......but it takes time....
                        solution:
                        suppose there are 'n' teams playing in the competetion
                        name the team as x1 , x2 , x3 ...........xn
                        x1 will play n-1 games
                        x2 will also play n-1 games but we have already included match between x1 and x2 above.so it n-2
                        x3    ....... n-3
                        x4 ......... n-4
                        ...
                        ...
                        xn .......n-n
                        now add them
                        (n-1) + (n-2) + (n-3)........(n-n) = 45   (given 45)
                        (n+n+n......n times) - (1 + 2 + 3 + 4.......n) =45
                        n^2 - (n(n+1)/2) = 45
                        2*n^2 - n^2 - n = 90
                        n^2 - n -90 = 0
                        (n - 10)*(n + 9) = 0
                        therefore n = 10    (because n cannot be negative)
                         
                        is it k now???
                         


                        suvarchalarani <raninistala@...> wrote:
                        2.in a competition a several teams take part. each
                        > team plays against all opp team. total no games played
                        > is 45. how many teams are there?
                        > ans given is 10
                        > bt hw
                        ?


                        Hi Mandar

                        If u got the explanation to this as to how you arrive at nC2 then, fine.

                        Otherwise, think on these lines

                        A game could be played between any 2 teams, right?
                        when there are n teams , everytime 2 teams are selected in nC2 ways

                        given that 45 games are played
                        so, nC2=45

                        If you know by experience, that nC2=45 when n=10, good.

                        Otherwise, nC2=n(n-1)(n-2)!/(n-2)!2!

                        =>n(n-1)=45*2=90
                        =>n(n-1)=10*9
                        =>n=10

                        Or by lenghty method, solve the quadratic equation that results from
                        the above as:
                        n^2-n-90=0

                        Discard the negative value for n and consider the positive value only

                        Hope this explanation helped you

                        Good luck



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