17M.1.AHL.TZ1.H_4

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

[N/A]