Date | May 2012 | Marks available | 3 | Reference code | 12M.2.hl.TZ2.4 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | How many | Question number | 4 | Adapted from | N/A |
Question
Fifteen boys and ten girls sit in a single line.
In how many ways can they be seated in a single line so that the boys and girls are in two separate groups?
Two boys and three girls are selected to go the theatre. In how many ways can this selection be made?
Markscheme
number of arrangements of boys is \(15!\) and number of arrangements of girls is \(10!\) (A1)
total number of arrangements is \(15! \times 10! \times 2( = 9.49 \times {10^{18}})\) M1A1
Note: If 2 is omitted, award (A1)M1A0.
[3 marks]
number of ways of choosing two boys is \(\left( {\begin{array}{*{20}{c}}
{15} \\
2
\end{array}} \right)\) and the number of ways of choosing three girls is \(\left( {\begin{array}{*{20}{c}}
{10} \\
3
\end{array}} \right)\) (A1)
number of ways of choosing two boys and three girls is \(\left( {\begin{array}{*{20}{c}}
{15} \\
2
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
{10} \\
3
\end{array}} \right) = 12600\) M1A1
[3 marks]
Examiners report
A good number of correct answers were seen to this question, but a significant number of candidates forgot to multiply by 2 in part (a) and in part (b) the most common error was to add the combinations rather than multiply them.
A good number of correct answers were seen to this question, but a significant number of candidates forgot to multiply by 2 in part (a) and in part (b) the most common error was to add the combinations rather than multiply them.