Moderate -0.5 This is a standard permutations problem with a restriction. Students arrange 5 men first (5! ways), then place 2 women in the 6 gaps created (6P2 ways). The technique is well-established and commonly taught, requiring only straightforward application of the 'gaps method' with no novel insight, making it slightly easier than average.
\(\ldots M \ldots M \ldots M \ldots M \ldots M \ldots\)
M1
\(k \times 5!\) (120) or \(k \times {}^6P_2\) (30), k is an integer \(\geqslant 1\)
No. ways men placed \(\times\) No. ways women placed in gaps \(= 5! \times {}^6P_2\)
M1
Correct unsimplified expression
\(= 3600\)
A1
Correct answer
Method 2:
Answer
Marks
Guidance
Number with women together \(= 6! \times 2\) (1440); Total number of arrangements \(= 7!\) (5040)
M1
\(6! \times 2\) or \(7! - k\) seen, k is an integer \(\geqslant 1\)
Number with women not together \(= 7! - 6! \times 2\)
M1
Correct unsimplified expression
\(= 3600\)
A1
Correct answer
Total: 3 marks
## Question 1:
**Method 1:**
$\ldots M \ldots M \ldots M \ldots M \ldots M \ldots$ | M1 | $k \times 5!$ (120) or $k \times {}^6P_2$ (30), k is an integer $\geqslant 1$
No. ways men placed $\times$ No. ways women placed in gaps $= 5! \times {}^6P_2$ | M1 | Correct unsimplified expression
$= 3600$ | A1 | Correct answer
**Method 2:**
Number with women together $= 6! \times 2$ (1440); Total number of arrangements $= 7!$ (5040) | M1 | $6! \times 2$ or $7! - k$ seen, k is an integer $\geqslant 1$
Number with women not together $= 7! - 6! \times 2$ | M1 | Correct unsimplified expression
$= 3600$ | A1 | Correct answer
**Total: 3 marks**
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1 A group consists of 5 men and 2 women. Find the number of different ways that the group can stand in a line if the women are not next to each other.\\
\hfill \mbox{\textit{CAIE S1 2018 Q1 [3]}}