April Bush

Answered

2022-06-25

Number of ways 10 men and 4 women can sit at a round table so there is no block of 3 consecutive women

In 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 $(}\genfrac{}{}{0ex}{}{10}{4}{\textstyle )$ 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 $(}\genfrac{}{}{0ex}{}{10}{3}{\textstyle )$ 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 $(}\genfrac{}{}{0ex}{}{10}{2}{\textstyle )$ ways and again, if I'm not wrong, place them in 4! ways.

Therefore, my answer to the number of possible ways is:

$9!\cdot 4!({\textstyle (}\genfrac{}{}{0ex}{}{10}{4}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{10}{3}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{10}{2}{\textstyle )}).$.

In 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 $(}\genfrac{}{}{0ex}{}{10}{4}{\textstyle )$ 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 $(}\genfrac{}{}{0ex}{}{10}{3}{\textstyle )$ 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 $(}\genfrac{}{}{0ex}{}{10}{2}{\textstyle )$ ways and again, if I'm not wrong, place them in 4! ways.

Therefore, my answer to the number of possible ways is:

$9!\cdot 4!({\textstyle (}\genfrac{}{}{0ex}{}{10}{4}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{10}{3}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{10}{2}{\textstyle )}).$.

Answer & Explanation

Donavan Scott

Expert

2022-06-26Added 22 answers

Step 1

Your answer is missing factor of 3 in the second term.

$9!\cdot 4!({\textstyle (}\genfrac{}{}{0ex}{}{10}{4}{\textstyle )}+{3}\cdot {\textstyle (}\genfrac{}{}{0ex}{}{10}{3}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{10}{2}{\textstyle )})$

Why 3? That's because when you make groups of 2,1 and 1 women and choose 3 spaces between men to seat them which is $(}\genfrac{}{}{0ex}{}{10}{3}{\textstyle )$, you also have 3 ways to choose which of those 3 spaces will have two women.

Step 2

Alternatively, subtract arrangements which have 3 or 4 consecutive women from total arrangements of 13!. So the answer can also be written as, $13!-(10!\cdot 9+10!)\cdot 4!$

Explanation:

First term being subtracted - three consecutive women. Let'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's first seat all women together and then men can be arranged in 10! ways. Finally women can be arranged within in 4! ways.

Your answer is missing factor of 3 in the second term.

$9!\cdot 4!({\textstyle (}\genfrac{}{}{0ex}{}{10}{4}{\textstyle )}+{3}\cdot {\textstyle (}\genfrac{}{}{0ex}{}{10}{3}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{10}{2}{\textstyle )})$

Why 3? That's because when you make groups of 2,1 and 1 women and choose 3 spaces between men to seat them which is $(}\genfrac{}{}{0ex}{}{10}{3}{\textstyle )$, you also have 3 ways to choose which of those 3 spaces will have two women.

Step 2

Alternatively, subtract arrangements which have 3 or 4 consecutive women from total arrangements of 13!. So the answer can also be written as, $13!-(10!\cdot 9+10!)\cdot 4!$

Explanation:

First term being subtracted - three consecutive women. Let'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'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

Expert

2022-06-27Added 9 answers

Step 1

Let'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'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 2

For the first case, there are $(}\genfrac{}{}{0ex}{}{4}{3}{\textstyle )}\cdot 3!$ 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, $10\cdot 2\cdot 9!$ of these have a block of 4 women. So there are a total of $216\cdot 10!$ ways to arrange this case.

Step 3

For 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 $24\cdot 10!$

So the total is $13!-240\cdot 10!=\overline{){\displaystyle 1476\cdot 10!}}$

Let'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'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 2

For the first case, there are $(}\genfrac{}{}{0ex}{}{4}{3}{\textstyle )}\cdot 3!$ 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, $10\cdot 2\cdot 9!$ of these have a block of 4 women. So there are a total of $216\cdot 10!$ ways to arrange this case.

Step 3

For 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 $24\cdot 10!$

So the total is $13!-240\cdot 10!=\overline{){\displaystyle 1476\cdot 10!}}$

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