Number of ways 10 men and 4 women can sit at a round table so there is no block of 3 consecutive womenIn how many ways can 10 men and 4 women sit at a round table so that there is no block of 3 consecutive women?Since there is no restriction on men, I thought they should sit first. Since the table is round, I fixed one seat for 1 in 10 men and placed the rest of them 9! ways and then we have 10 gaps in which we can place 4 women, at most 2 in each. I think these are the possibilities:1. One woman in each gap: we can choose those gaps in ( 10 4 ) ways and then place women in 4! different ways.2. 2 women in 1 gap and the remaining 2 in each of their own. We need 3 gaps altogether, which can be chosen in ( 10 3 ) ways. Since women are different entities, I believe we should be able to place them in, again 4! ways.3. 2 women in 2 gaps: 2 in each. We can choose 2 gaps in ( 10 2 ) ways and again, if I'm not wrong, place them in 4! ways.Therefore, my answer to the number of possible ways is: 9 ! ⋅ 4 ! ( ( 10 4 ) + ( 10 3 ) + ( 10 2 ) ) ..

Question

Number of ways 10 men and 4 women can sit at a round table so there is no block of 3 consecutive womenIn how many ways can 10 men and 4 women sit at a round table so that there is no block of 3 consecutive women?Since there is no restriction on men, I thought they should sit first. Since the table is round, I fixed one seat for 1 in 10 men and placed the rest of them 9! ways and then we have 10 gaps in which we can place 4 women, at most 2 in each. I think these are the possibilities:1. One woman in each gap: we can choose those gaps in                     (                    10      4                      )             ways and then place women in 4! different ways.2. 2 women in 1 gap and the remaining 2 in each of their own. We need 3 gaps altogether, which can be chosen in                     (                    10      3                      )             ways. Since women are different entities, I believe we should be able to place them in, again 4! ways.3. 2 women in 2 gaps: 2 in each. We can choose 2 gaps in                     (                    10      2                      )             ways and again, if I&#039;m not wrong, place them in 4! ways.Therefore, my answer to the number of possible ways is:  9  !  ⋅  4  !      (                            (                            10        4                              )                      +                            (                            10        3                              )                      +                            (                            10        2                              )                      )    ..

Donavan Scott · Accepted Answer

Step 1Your answer is missing factor of 3 in the second term.      9    !    ⋅    4    !          (                                    (                                    10          4                                      )                              +              3            ⋅                                    (                                    10          3                                      )                              +                                    (                                    10          2                                      )                              )      Why 3? That&#039;s because when you make groups of 2,1 and 1 women and choose 3 spaces between men to seat them which is                               (                            10        3                              )                    , you also have 3 ways to choose which of those 3 spaces will have two women.Step 2Alternatively, subtract arrangements which have 3 or 4 consecutive women from total arrangements of 13!. So the answer can also be written as,       13    !    −    (    10    !    ⋅    9    +    10    !    )    ⋅    4    !  Explanation:First term being subtracted - three consecutive women. Let&#039;s first seat three women together and then men can be arranged in 10! ways. The lone woman can sit in 9 spaces between men. Finally women can be arranged within in 4! ways.Second term being subtracted - all four women sit together. Let&#039;s first seat all women together and then men can be arranged in 10! ways. Finally women can be arranged within in 4! ways.

Poftethef9t · Accepted Answer

Step 1Let&#039;s say the seats are numbered 1 to 14. Fix an arbitrary man at seat 1. Now we have 9 men left and 4 women left to seat.Now consider the remaining 13 seats. Since the man in seat 1 is blocking any possibility of a consecutive block of women wrapping around, we just need to find the number of ways to put the remaining 13 people in a line so that we don&#039;t have a consecutive block of 3 women.We will use complimentary counting. There are of course 13! ways to seat the 13 people with no attention to restrictions. The only way to have a consecutive block of 3 women is if there is a block of exactly 3 women or a block of exactly 4 women.Step 2For the first case, there are                     (                    4      3                      )              ⋅  3  ! ways to choose the 3 women in the block and then order them. There are 11! ways to arrange the   9  +  2  =  11 entities. However,   10  ⋅  2  ⋅  9  ! of these have a block of 4 women. So there are a total of   216  ⋅  10  ! ways to arrange this case.Step 3For the second case, there are 4! ways to arrange the 4 women in the block. There are 10! ways to arrange the 10 entities. Hence the total is   24  ⋅  10  !So the total is   13  !  −  240  ⋅  10  !  =                              1476          ⋅          10          !