There are 4 girls and 4 boys including John and Martin. In how many ways can they be arranged such that two girls will be behind John and Martin? 
There are 4 girls and 4 boys including John and Martin. In how many
  ways can they be arranged such that two girls will be behind John and
  Martin?

There are $2!$ ways to permutate John and Martin. As two girls need to be behind John and Martin, we have that
$$GGBBJMGG$$
Now there are $10$ ways to pick girls and $2!$ ways to permutate boys. Could you assist?
Regards
 A: 
There are four boys and four girls, including John and Martin.  In how many ways can they be arranged so that there are exactly two girls behind John and Martin?

Choose six of the eight positions for John, Martin, and the two girls, which can be done in $\binom{8}{6}$ ways.  If there are exactly two girls behind John and Martin, then John and Martin must be in the middle two of the six selected positions.  They can be arranged in those positions in $2!$ ways.  The four girls can be arranged in the remaining four selected positions in $4!$ ways.  The remaining two boys can be arranged in the remaining two positions in $2!$ ways.  Hence, the number of arrangements in which there are exactly two girls behind John and Martin is 
$$\binom{8}{6}2!4!2!$$

There are four boys and four girls, including John and Martin.  In how many ways can they be arranged so that there are at least two girls behind John and Martin?

Choose six of the eight positions for John, Martin, and the two girls, which can be done in $\binom{8}{6}$ ways.  Since John and Martin must have at least two girls behind them, they must be in the first four of those six selected positions.  Thus, among those six positions, there are $\binom{4}{2}$ ways to select the positions of John and Martin.  John and Martin may be arranged in those two positions in $2!$ ways.  The four girls may be arranged in the remaining four of the original six selected positions in $4!$ ways.  The remaining two boys may be arranged in the remaining two positions in $2!$ ways.  Consequently, there are 
$$\binom{8}{6}\binom{4}{2}2!4!2!$$
arrangements in which there are at least two girls behind John and Martin. 
