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 ?
closed as off-topic by Namaste, Adam Hughes, Leucippus, JMoravitz, user228113 Jan 16 '17 at 19:30
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$13!$ 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 $=13!-10!×4!$
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.
Calculate the total number of ways of seating everyone. Then subtract how many ways distinct ways in which 4 girls may sit together.
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 = $13! - 10! \times 4!$