Three girls and four boys are seated randomly on a straight bench. Find the probability that the girls sit together and the boys sit together.

METHOD 1

total number of arrangements 7!     (A1)

number of ways for girls and boys to sit together $= 3! \times 4! \times 2$     (M1)(A1)

Note:    Award M1A0 if the 2 is missing.

probability $\frac{{3! \times 4! \times 2}}{{7!}}$     M1

Note:     Award M1 for attempting to write as a probability.

$\frac{{2 \times 3 \times 4! \times 2}}{{7 \times 6 \times 5 \times 4!}}$

$= \frac{2}{{35}}$     A1

Note:     Award A0 if not fully simplified.

METHOD 2

$\frac{3}{7} \times \frac{2}{6} \times \frac{1}{5} + \frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4}$     (M1)A1A1

Note:     Accept $\frac{3}{7} \times \frac{2}{6} \times \frac{1}{5} \times 2$ or $\frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} \times 2$.

$= \frac{2}{{35}}$     (M1)A1

Note:     Award A0 if not fully simplified.

[5 marks]