3 girls and 4 boys were standing in a circle . What is the probability that two girls are together but one is not with them?

Question

Question: 3 girls and 4 boys were standing in a circle . What is the probability that two girls are together but one is not with them ?

Answer 1

a bit off topic but are you applying to Oxbridge?

And now to the question;

The total # ways of arranging is$\frac{7!}{7}$ (due to rotations)

The # of ways of putting two girls together is $3\choose 2$ Now they must have boys on either side so $4 \choose 2$ And finally the other 3 can be arranged in $3! ways$

Pick two adjacent slots from your 7 $6\choose 1$ then put the two girls in and multiply by 2 as they can be arranged in 2 ways Then select the adjacent boys and again multiply by 2 and finally arrange the last.(account for rotations) This gives; $$\frac{{6\choose 1}\cdot {3\choose 2}\cdot 2 \cdot {4\choose 2}\cdot 2 \cdot 3!}{6}$$ ways and hence the probability is the second expression divided by the first.

Answer 2

Since only probability has been asked for, there is a very simple way.

Ignore the boys, and seat any two girls together.
The $3rd$ girl has $3$ permissible places out of 5,
thus $Pr = \dfrac35 = 0.6$

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If you insist on doing it in a conventional manner:

Unless otherwise specified, seats in a circle are taken as unnumbered, and formula is $(n-1)!$

Thus total ways of seating = $6!$

For the favorable ways :-
$\binom32 = 3$ ways of choosing the two "together" girls,
$2$ ways of seating them somewhere $(A-B / B-A)$,
$3$ ways of seating the remaining girl non-adjacently,
$4!$ ways of seating the boys,

Putting everything together, $Pr = \dfrac{3\cdot2\cdot3\cdot4!}{6!} = 0.6$