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If $5$ boys and $5$ girls arranged in a circle alternatively. If a perticular boy and girl are adjacent to each other in any arrangement , is

solution i try

number of ways in which boys arrange in a circle is $(5-1)!$

and arrange $5$ boys in these $5$ gap is $5!$

so total possibilities in which boys and girl sit alternatively is $4!\times 5!$

but answer given us different from my answer. help me

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  • $\begingroup$ Should the question say how many seating arrangements are there in which five boys and five girls sit alternately in which a particular boy and girl are adjacent to each other? $\endgroup$ – N. F. Taussig May 4 '18 at 21:25
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If two arrangements are considered identical if they differ only by a rotations, I would say, that it doesn't matter where the particular boy sits. Then the other boys may be seated in $4!$ ways. There are two chairs next to the particular boy where the particular girl mays sit, and then the remaining girls can be seated in $4!$ ways, giving $$2\cdot4!\cdot4!$$ ways in all.

If two arrangements are the same if they differ by a reflection, (that is they are the same except that in once case the children are arranged clockwise and counterclockwise in the other) then you have to divide by $2$.

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let $B_{1},B_{2}....B_{5}$ be $5$ boys and $G_{1},G_{2},......G_{5}$ be $5$ girls

let's say particular boy $B_{1}$ has to sit adjacent to $G_{1}$ in all arrangements so tie them togeather and treat them as a single block now you have $1block + 4 boys +4girls$ left to arrange

seat $1block+4boys $ in $4!$ ways around the table now there are $5$ gaps in between them but girls can't sit beside the girl in the block(because girls and boys has to sit alternately) so 4 leftover girls can occupy only $4 $gaps in $4!$ ways .

now use multiplication principle to get answer as $4!.4!=(4!)^2$ ways to seat all of them alternately with one particular boy and one particular girl together

edit :

i'm here considering two arrangements identical if they differ by reflection, if you consider reflections as different arrangements then you have to multiply my answer by $2$ i.e, $2.(4!)^2$ ways

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