Arrangement of 12 boys and 2 girls in a row.

Question

12 boys and 2 girls in a row are to be seated in such a way that at least 3 boys are present between the 2 girls. My result: Total number of arrangements = 14! P₁ = number of ways girls can sit together = $2!×13!$
Now I want to find P₂ the number of ways in one boy sits between the two girls and then P₃ the number of ways in which two boys sit between the two girls. How to find these two?

Answer 1

The number of arrangements in which exactly one boy sits between the girls is $$12 \cdot 2! \cdot 12!$$ since there are twelve ways to choose the boy who sits between the girls, two ways of choosing the girl who sits to his left, one way of choosing the girl who sits to his right, and $12!$ ways of arranging the block of three people and the other eleven boys.

The number of arrangements in which exactly two boys sit between the girls is $$12 \cdot 11 \cdot 2! \cdot 11!$$ since there are twelve ways to choose the boy who sits in the first seat between the two girls, eleven ways to choose the boy who sits in the second seat between the two girls, two ways to choose the girl who sits to their left, one way of choosing the girl who sits to their right, and $11!$ ways to arrange the block of four people and the other ten boys.

Notice that $$14! - 2!13! - 12 \cdot 2!12! - 12 \cdot 11 \cdot 2!11! = \binom{11}{2}2!12!$$ in agreement with the answers provided by drhab and Henning Makholm.

Answer 2

Do it like you did for only the girls: find out how many ways there are of arranging the block consisting of the girls plus the boys between them, then how many ways there are of placing that block among the rest of the boys.

Answer 3

P2) Left most girl can be at any of 12 positions. (i.e. she can't be the two right most positions). Among the 12 boys, they can be arranged in any way (12!). Also, the two girls can switch (2!).

So 12!*2!*12

P3)Left most girl can be at any of 11 positions. Otherwise same analysis. So: 12!*2!*11

Answer 4

With stars and bars we find that there are $\binom{9+2}{2}=\binom{11}{2}=55$ different summations $a+b+c=9$ where $a,b,c$ are nonnegative integers.

So there are also $55$ different summations $a+b+c=12$ under the extra condition that $b\geq3$.

Here $b$ corresponds with the number of boys sitting in between the girls.

There are $12!$ arrangements for the boys and $2!$ for the girls, so we end up with:

$$55\times12!\times2!$$ posssibilities.

Answer 5

First arrange nine Bs and two Gs in any order you want, which can be done in $\binom{11}2$ ways. Then add three more Bs between the two Gs.

Finally multiply by $2!\cdot 12!$ if you want to count lists of names rather than just sequences of genders.

Answer 6

Pick the two positions the girls are in ($12$ for the first part, $11$ for the second part) and then choose which girl is on the left ($2$ choices). Arrange the boys however you like ($12!$)..

So, $P_2 = 12 \times 2 \times 12!$, and $P_3 = 11 \times 2 \times 12!$.