| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Selection from categorized items |
| Difficulty | Moderate -0.8 This question tests standard combinations with simple restrictions. Part (a) requires basic counting with fixed categories (3C1 × 5C1 for (i), 5C1 × 6C2 for (ii)), and part (b) involves routine permutations with repeated letters (9!/2!2! and 5!/2! respectively). All techniques are direct applications of formulas with no problem-solving insight required, making it easier than average but not trivial due to multiple parts. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For multiplying two combinations together | M1 | |
| For correct answer | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For seeing \(_6C_2\), or separating it into three alternatives either added or multiplied | M1 | |
| For correct answer | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For dividing by 2! twice | M1 | |
| For correct answer | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 5! seen in a numerator | B1 | |
| For correct final answer | B1 [2] |
(a)(i) $_3C_1 \times {}_5C_1 = 15$
| Answer | Marks | Guidance |
|--------|-------|----------|
| For multiplying two combinations together | M1 | |
| For correct answer | B1 [2] | |
(ii) $_5C_1 \times {}_6C_2 = 75$
| Answer | Marks | Guidance |
|--------|-------|----------|
| For seeing $_6C_2$, or separating it into three alternatives either added or multiplied | M1 | |
| For correct answer | A1 [2] | |
(b)(i) $9!/2!2! = 90720$
| Answer | Marks | Guidance |
|--------|-------|----------|
| For dividing by 2! twice | M1 | |
| For correct answer | A1 [2] | |
(ii) $5!$ Or $_5P_5 = 120$
| Answer | Marks | Guidance |
|--------|-------|----------|
| 5! seen in a numerator | B1 | |
| For correct final answer | B1 [2] | |7
\begin{enumerate}[label=(\alph*)]
\item A football team consists of 3 players who play in a defence position, 3 players who play in a midfield position and 5 players who play in a forward position. Three players are chosen to collect a gold medal for the team. Find in how many ways this can be done
\begin{enumerate}[label=(\roman*)]
\item if the captain, who is a midfield player, must be included, together with one defence and one forward player,
\item if exactly one forward player must be included, together with any two others.
\end{enumerate}\item Find how many different arrangements there are of the nine letters in the words GOLD MEDAL
\begin{enumerate}[label=(\roman*)]
\item if there are no restrictions on the order of the letters,
\item if the two letters D come first and the two letters L come last.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2005 Q7 [8]}}