1
$\begingroup$

Number of ways in which 5 boys and 4 girls can be seated around a circular table such that no two girls sit together and two particular boys are always together ?

The answer to this question is $3!2!4!$ . It is done by considering $2$ boys as one unit and the the number of units (of boys) is $4$ so they can be arranged in $3!2!$ ways. Then number of girls can be arranged in $4!$ ways. However , I have a doubt. After five boys have been seated aren’t there $5$ places created for girls ? So girls should actually be seated in $\binom{5}{4}4!$ ways right ? So final answer should be $3!2!\binom{5}{4}4!$ right ?

|cite|improve this question
$\endgroup$
  • $\begingroup$ There are 4 boys(2 boys are a group) so only 4 gaps are made. $\endgroup$ – prog_SAHIL Dec 14 '17 at 4:06
  • $\begingroup$ @prog_SAHIL oh yeah thanks . I totally missed that $\endgroup$ – Aditi Dec 14 '17 at 4:08
  • $\begingroup$ @prog_SAHIL yes , but remember ! It’s a circular permutation ! So if we have $n$ units then they are arranged in $(n-1)!$ ways . $\endgroup$ – Aditi Dec 14 '17 at 4:22
  • $\begingroup$ Are you preparing for JEE's? $\endgroup$ – prog_SAHIL Dec 14 '17 at 4:27
  • $\begingroup$ @prog_SAHIL yes! $\endgroup$ – Aditi Dec 14 '17 at 5:01
1
$\begingroup$

First seat the $4$ girls around the table, which can be done in $(4-1)!$ ways (why?, Check "circular permutation"). Now in between these girls, there are $4$ gaps, where you would have to seat the $5$ boys, two of whom must sit together. So we club the two boys to form a single unit and see that the $4$ units can be seated in the $4$ gaps in $4!$ ways. However the two boys who form the unit, can be rearranged among themselves in $2$ ways. So total number of ways is $2!\times3!\times4!$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Okay I understood ! Thanks a lot ! $\endgroup$ – Aditi Dec 14 '17 at 4:26
1
$\begingroup$

Considering 2 boys as a unit. (I will treat them as 4 boys)

Now there are 4 boys, around a circular table only 4 gaps will be made.(5 if it was a straight queue)

$4!$ Permutating 4 boys

$2!$ Internal permutation of 2 boys in the group

$3!$ Permutating 4 girls

$$3!2!4!$$

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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