-3
$\begingroup$

In how many ways 10 boys and 4 girls can be seated at a round table such that all the 4 girls do not sit together ?

$\endgroup$

closed as off-topic by Namaste, Adam Hughes, Leucippus, JMoravitz, user228113 Jan 16 '17 at 19:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Adam Hughes, Leucippus, JMoravitz
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ i am not getting this .....all the 4 girls do not sit together $\endgroup$ – jerry Jan 16 '17 at 18:10
  • $\begingroup$ It depends on your definition of the table. Are the boys and girls distinct (I hope they are...)? Are all rotations of the table the same? $\endgroup$ – pie314271 Jan 16 '17 at 18:12
  • 1
    $\begingroup$ by "all the 4 girls do not sit together" do you mean that you may not have four girls in a row but three girls in a row and the other girl separated from the others is okay? or do you mean that no two girls may sit next to one another? $\endgroup$ – JMoravitz Jan 16 '17 at 18:13
  • $\begingroup$ @JMoravitz i wanted to know both the conditions ..but three girls in a row and the other girl separated and no two girls may sit next to one another $\endgroup$ – jerry Jan 16 '17 at 18:33
  • $\begingroup$ @jerry: look over your question, the answers to it (one of which has been deleted, but still put more thought in than you did to your question), with comments from others, and then review all the comments. What work have you done, or are doing, compared to all of the others trying to help you? $\endgroup$ – Namaste Jan 16 '17 at 18:50
2
$\begingroup$

$13!$ ways to arrange the $14$ persons around a round table

group the $4$ girls and consider as a single girl. then total $11$ persons and they can be arranged in $10!$ ways. $4!$ ways of arranging the girls among themselves.

Required ways $=13!-10!×4!$

$\endgroup$
  • 2
    $\begingroup$ wait oops I was wrong with 13!-4!, but shouldn't this be 13!-4!*10!? Because you have to permutate the girls $\endgroup$ – pie314271 Jan 16 '17 at 18:17
  • $\begingroup$ yes, you are right. i missed it. editing my answer $\endgroup$ – Kiran Jan 16 '17 at 18:18
  • 1
    $\begingroup$ As an aside, with the correction made this answers the question assuming the answer to my question asking for clarification is "yes" that it is allowed that 3 girls sit next to one another so long as the fourth is separated. In the event that the OP gives clarification and specifies otherwise that no girl is allowed to sit next to any other girl then this will no longer be correct. $\endgroup$ – JMoravitz Jan 16 '17 at 18:20
0
$\begingroup$

For the question of if no two girls may sit next to eachother:

Begin with a circular table with no chairs

  • Start by putting the youngest girl with a chair at the table in an arbitrary position. (does not matter where, does not affect count, this is just so that we have a point of reference)

  • Add ten chairs to the table and clockwise from where the girl is sitting, seat a boy at each chair.

  • Choose three of the positions between two boys to add an additional chair.

  • Choose which remaining girl sits in which of the newly added chairs clockwise from the youngest girl's perspective.

Figure out how many options there are at each step and apply multiplication principle.

For the question of if two girls may sit next to eachother so long as all four of the girls aren't in a row, see @Kiran's answer.

$\endgroup$
0
$\begingroup$

Hint -

Calculate the total number of ways of seating everyone. Then subtract how many ways distinct ways in which 4 girls may sit together.

Edit -

In a round arrangement we first have to fix the position for the first person, which can be performed in only one way. Once we have fixed the position for the first person** we can now arrange the remaining $(n-1)$ persons in $(n-1)!$ ways.

So total ways = 13!

Now treat all 4 girls as 1 group so we have 9 boys + 1 group = 10 people so they can be arrange in 10!.

But girls in group further can sit in 4! ways.

So we have $10! \times 4!$

Number of ways 4 girls not sit together = $13! - 10! \times 4!$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.