CAIE S1 2016 November — Question 1 3 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2016
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCombinations & Selection
TypeSelection with family/relationship restrictions
DifficultyModerate -0.3 This is a straightforward combinations problem with a simple restriction. Students need to recognize the complementary counting approach (total committees minus committees with both) or use case-by-case counting (committees with William only, Mary only, or neither). The calculation involves basic combination formulas C(n,r) with small numbers and simple arithmetic—standard fare for S1 with no conceptual surprises.
Spec5.01a Permutations and combinations: evaluate probabilities
1 A committee of 5 people is to be chosen from 4 men and 6 women. William is one of the 4 men and Mary is one of the 6 women. Find the number of different committees that can be chosen if William and Mary refuse to be on the committee together.
Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Total ways \(^{10}C_5 = 252\)M1 \(^{10}C_5 - \ldots\) or \(252 - \ldots\)
MW together e.g. \((MW)^{***}\) in \(^8C_3\) ways \(= 56\)B1 252 and 56 seen, may be unsimplified
MW not together \(= 252 - 56 = 196\) waysA1 [3]
OR 1: \(2\cdot{}^8C_4 + {}^8C_5\)M1 \(2\cdot{}^nC_4 + {}^nC_5\)
\(2\cdot{}^8C_4 = 2\times70 = 140\); \(^8C_5 = 56\)B1 140 and 56 seen, may be unsimplified
\(2\cdot{}^8C_4 + {}^8C_5 = 196\)A1
OR 2: \(2\cdot{}^9C_5 - {}^8C_5\)M1 \(2\cdot{}^9C_5 - \ldots\)
\(2\cdot{}^9C_5 = 2\times126 = 252\); \(^8C_5 = 56\)B1 252 and 56 seen, may be unsimplified
\(2\cdot{}^9C_5 - {}^8C_5 = 196\)A1 
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Total ways $^{10}C_5 = 252$ | M1 | $^{10}C_5 - \ldots$ or $252 - \ldots$ |
| MW together e.g. $(MW)^{***}$ in $^8C_3$ ways $= 56$ | B1 | 252 and 56 seen, may be unsimplified |
| MW not together $= 252 - 56 = 196$ ways | A1 [3] | |
| **OR 1:** $2\cdot{}^8C_4 + {}^8C_5$ | M1 | $2\cdot{}^nC_4 + {}^nC_5$ |
| $2\cdot{}^8C_4 = 2\times70 = 140$; $^8C_5 = 56$ | B1 | 140 and 56 seen, may be unsimplified |
| $2\cdot{}^8C_4 + {}^8C_5 = 196$ | A1 | |
| **OR 2:** $2\cdot{}^9C_5 - {}^8C_5$ | M1 | $2\cdot{}^9C_5 - \ldots$ |
| $2\cdot{}^9C_5 = 2\times126 = 252$; $^8C_5 = 56$ | B1 | 252 and 56 seen, may be unsimplified |
| $2\cdot{}^9C_5 - {}^8C_5 = 196$ | A1 | |

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1 A committee of 5 people is to be chosen from 4 men and 6 women. William is one of the 4 men and Mary is one of the 6 women. Find the number of different committees that can be chosen if William and Mary refuse to be on the committee together.

\hfill \mbox{\textit{CAIE S1 2016 Q1 [3]}}
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