number of ways $3$ teachers can be invited for a guest lecturer on $6$ days in a school












0












$begingroup$



In how many ways $3$ teachers can be invited for a guest lecturer on $6$ days in a school so that every teacher is invited at least once (on any given day exactly one teacher is invited)




Try: First Teacher $T_{1}$ has $6$ possiability (In any one of $6$ days)



Second Teacher $T_{2}$ has $5$ Possiability



Third Teacher $T_{3}$ has $4$ possiability



So total $6times 5 times 4 = 120$



But answer given as $540$



Could some help me how to count total ways so the answer is $540$



Thanks in advance






combinatorics







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  • 3




    $begingroup$
    You are only filling 3 days. You need a guest on all 6 days. So, some teachers will need to repeat. There does not appear to be a further restriction so it could be that all visit twice or one comes four times and the others just once.
    $endgroup$
    – badjohn
    Jan 9 at 9:08
















0












$begingroup$



In how many ways $3$ teachers can be invited for a guest lecturer on $6$ days in a school so that every teacher is invited at least once (on any given day exactly one teacher is invited)




Try: First Teacher $T_{1}$ has $6$ possiability (In any one of $6$ days)



Second Teacher $T_{2}$ has $5$ Possiability



Third Teacher $T_{3}$ has $4$ possiability



So total $6times 5 times 4 = 120$



But answer given as $540$



Could some help me how to count total ways so the answer is $540$



Thanks in advance






combinatorics







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    You are only filling 3 days. You need a guest on all 6 days. So, some teachers will need to repeat. There does not appear to be a further restriction so it could be that all visit twice or one comes four times and the others just once.
    $endgroup$
    – badjohn
    Jan 9 at 9:08














0












0








0





$begingroup$



In how many ways $3$ teachers can be invited for a guest lecturer on $6$ days in a school so that every teacher is invited at least once (on any given day exactly one teacher is invited)




Try: First Teacher $T_{1}$ has $6$ possiability (In any one of $6$ days)



Second Teacher $T_{2}$ has $5$ Possiability



Third Teacher $T_{3}$ has $4$ possiability



So total $6times 5 times 4 = 120$



But answer given as $540$



Could some help me how to count total ways so the answer is $540$



Thanks in advance






combinatorics







share|cite|improve this question











$endgroup$





In how many ways $3$ teachers can be invited for a guest lecturer on $6$ days in a school so that every teacher is invited at least once (on any given day exactly one teacher is invited)




Try: First Teacher $T_{1}$ has $6$ possiability (In any one of $6$ days)



Second Teacher $T_{2}$ has $5$ Possiability



Third Teacher $T_{3}$ has $4$ possiability



So total $6times 5 times 4 = 120$



But answer given as $540$



Could some help me how to count total ways so the answer is $540$



Thanks in advance






combinatorics




combinatorics






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edited Jan 9 at 17:06









user

4,0001627




4,0001627










asked Jan 9 at 9:03









DXTDXT

5,6742630




5,6742630








  • 3




    $begingroup$
    You are only filling 3 days. You need a guest on all 6 days. So, some teachers will need to repeat. There does not appear to be a further restriction so it could be that all visit twice or one comes four times and the others just once.
    $endgroup$
    – badjohn
    Jan 9 at 9:08














  • 3




    $begingroup$
    You are only filling 3 days. You need a guest on all 6 days. So, some teachers will need to repeat. There does not appear to be a further restriction so it could be that all visit twice or one comes four times and the others just once.
    $endgroup$
    – badjohn
    Jan 9 at 9:08








3




3




$begingroup$
You are only filling 3 days. You need a guest on all 6 days. So, some teachers will need to repeat. There does not appear to be a further restriction so it could be that all visit twice or one comes four times and the others just once.
$endgroup$
– badjohn
Jan 9 at 9:08




$begingroup$
You are only filling 3 days. You need a guest on all 6 days. So, some teachers will need to repeat. There does not appear to be a further restriction so it could be that all visit twice or one comes four times and the others just once.
$endgroup$
– badjohn
Jan 9 at 9:08










2 Answers
2






active

oldest

votes


















2












$begingroup$

Undesired events:




  • A single teacher teaches repeately. $3$ ways to do so.

  • Only two teachers. First choose one of the teacher to exclude him/her, and then we avoid two extreme cases where one teacher teaches everyday.


$$3cdot (2^6-2)$$



Hence the undesired outcomes are $$3(2^6-1)$$



Without restriction, we could have $3^6$. Hence



$$3^6-3(2^6-1)=3(3^5-2^6+1)=3(243-64+1)=3(180)=540$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The table shows the teacher, the number of visits and the number of ways:
    $$begin{array}{c|c|c|c}
    T_1&T_2&T_3&text{No of visits}\
    hline
    1&1&4&{6choose 1}{5choose 1}=color{red}{30}\
    1&2&3&{6choose 1}{5choose 2}=color{green}{60}\
    1&3&2&{6choose 1}{5choose 3}=color{green}{60}\
    1&4&1&{6choose 1}{5choose 4}=color{red}{30}\
    2&1&3&{6choose 2}{4choose 1}=color{green}{60}\
    2&2&2&{6choose 2}{4choose 2}=color{blue}{90}\
    2&3&1&{6choose 2}{4choose 3}=color{green}{60}\
    3&1&2&{6choose 3}{3choose 1}=color{green}{60}\
    3&2&1&{6choose 3}{3choose 2}=color{green}{60}\
    4&1&1&{6choose 4}{2choose 1}=color{red}{30}\
    hline
    &&&color{red}{90}+color{green}{360}+color{blue}{90}=540.
    end{array}$$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

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      active

      oldest

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      2












      $begingroup$

      Undesired events:




      • A single teacher teaches repeately. $3$ ways to do so.

      • Only two teachers. First choose one of the teacher to exclude him/her, and then we avoid two extreme cases where one teacher teaches everyday.


      $$3cdot (2^6-2)$$



      Hence the undesired outcomes are $$3(2^6-1)$$



      Without restriction, we could have $3^6$. Hence



      $$3^6-3(2^6-1)=3(3^5-2^6+1)=3(243-64+1)=3(180)=540$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Undesired events:




        • A single teacher teaches repeately. $3$ ways to do so.

        • Only two teachers. First choose one of the teacher to exclude him/her, and then we avoid two extreme cases where one teacher teaches everyday.


        $$3cdot (2^6-2)$$



        Hence the undesired outcomes are $$3(2^6-1)$$



        Without restriction, we could have $3^6$. Hence



        $$3^6-3(2^6-1)=3(3^5-2^6+1)=3(243-64+1)=3(180)=540$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Undesired events:




          • A single teacher teaches repeately. $3$ ways to do so.

          • Only two teachers. First choose one of the teacher to exclude him/her, and then we avoid two extreme cases where one teacher teaches everyday.


          $$3cdot (2^6-2)$$



          Hence the undesired outcomes are $$3(2^6-1)$$



          Without restriction, we could have $3^6$. Hence



          $$3^6-3(2^6-1)=3(3^5-2^6+1)=3(243-64+1)=3(180)=540$$






          share|cite|improve this answer









          $endgroup$



          Undesired events:




          • A single teacher teaches repeately. $3$ ways to do so.

          • Only two teachers. First choose one of the teacher to exclude him/her, and then we avoid two extreme cases where one teacher teaches everyday.


          $$3cdot (2^6-2)$$



          Hence the undesired outcomes are $$3(2^6-1)$$



          Without restriction, we could have $3^6$. Hence



          $$3^6-3(2^6-1)=3(3^5-2^6+1)=3(243-64+1)=3(180)=540$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 9:08









          Siong Thye GohSiong Thye Goh

          101k1466117




          101k1466117























              1












              $begingroup$

              The table shows the teacher, the number of visits and the number of ways:
              $$begin{array}{c|c|c|c}
              T_1&T_2&T_3&text{No of visits}\
              hline
              1&1&4&{6choose 1}{5choose 1}=color{red}{30}\
              1&2&3&{6choose 1}{5choose 2}=color{green}{60}\
              1&3&2&{6choose 1}{5choose 3}=color{green}{60}\
              1&4&1&{6choose 1}{5choose 4}=color{red}{30}\
              2&1&3&{6choose 2}{4choose 1}=color{green}{60}\
              2&2&2&{6choose 2}{4choose 2}=color{blue}{90}\
              2&3&1&{6choose 2}{4choose 3}=color{green}{60}\
              3&1&2&{6choose 3}{3choose 1}=color{green}{60}\
              3&2&1&{6choose 3}{3choose 2}=color{green}{60}\
              4&1&1&{6choose 4}{2choose 1}=color{red}{30}\
              hline
              &&&color{red}{90}+color{green}{360}+color{blue}{90}=540.
              end{array}$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                The table shows the teacher, the number of visits and the number of ways:
                $$begin{array}{c|c|c|c}
                T_1&T_2&T_3&text{No of visits}\
                hline
                1&1&4&{6choose 1}{5choose 1}=color{red}{30}\
                1&2&3&{6choose 1}{5choose 2}=color{green}{60}\
                1&3&2&{6choose 1}{5choose 3}=color{green}{60}\
                1&4&1&{6choose 1}{5choose 4}=color{red}{30}\
                2&1&3&{6choose 2}{4choose 1}=color{green}{60}\
                2&2&2&{6choose 2}{4choose 2}=color{blue}{90}\
                2&3&1&{6choose 2}{4choose 3}=color{green}{60}\
                3&1&2&{6choose 3}{3choose 1}=color{green}{60}\
                3&2&1&{6choose 3}{3choose 2}=color{green}{60}\
                4&1&1&{6choose 4}{2choose 1}=color{red}{30}\
                hline
                &&&color{red}{90}+color{green}{360}+color{blue}{90}=540.
                end{array}$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The table shows the teacher, the number of visits and the number of ways:
                  $$begin{array}{c|c|c|c}
                  T_1&T_2&T_3&text{No of visits}\
                  hline
                  1&1&4&{6choose 1}{5choose 1}=color{red}{30}\
                  1&2&3&{6choose 1}{5choose 2}=color{green}{60}\
                  1&3&2&{6choose 1}{5choose 3}=color{green}{60}\
                  1&4&1&{6choose 1}{5choose 4}=color{red}{30}\
                  2&1&3&{6choose 2}{4choose 1}=color{green}{60}\
                  2&2&2&{6choose 2}{4choose 2}=color{blue}{90}\
                  2&3&1&{6choose 2}{4choose 3}=color{green}{60}\
                  3&1&2&{6choose 3}{3choose 1}=color{green}{60}\
                  3&2&1&{6choose 3}{3choose 2}=color{green}{60}\
                  4&1&1&{6choose 4}{2choose 1}=color{red}{30}\
                  hline
                  &&&color{red}{90}+color{green}{360}+color{blue}{90}=540.
                  end{array}$$






                  share|cite|improve this answer









                  $endgroup$



                  The table shows the teacher, the number of visits and the number of ways:
                  $$begin{array}{c|c|c|c}
                  T_1&T_2&T_3&text{No of visits}\
                  hline
                  1&1&4&{6choose 1}{5choose 1}=color{red}{30}\
                  1&2&3&{6choose 1}{5choose 2}=color{green}{60}\
                  1&3&2&{6choose 1}{5choose 3}=color{green}{60}\
                  1&4&1&{6choose 1}{5choose 4}=color{red}{30}\
                  2&1&3&{6choose 2}{4choose 1}=color{green}{60}\
                  2&2&2&{6choose 2}{4choose 2}=color{blue}{90}\
                  2&3&1&{6choose 2}{4choose 3}=color{green}{60}\
                  3&1&2&{6choose 3}{3choose 1}=color{green}{60}\
                  3&2&1&{6choose 3}{3choose 2}=color{green}{60}\
                  4&1&1&{6choose 4}{2choose 1}=color{red}{30}\
                  hline
                  &&&color{red}{90}+color{green}{360}+color{blue}{90}=540.
                  end{array}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 9 at 18:19









                  farruhotafarruhota

                  20.1k2738




                  20.1k2738






























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