CAIE S1 2023 March — Question 7 12 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2023
SessionMarch
Marks12
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Mark schemeDownload PDF ↗
TopicPermutations & Arrangements
TypeMultiple separation conditions combined
DifficultyStandard +0.3 This is a multi-part permutations question requiring systematic case-work and careful counting with repeated letters. Part (a) needs complementary counting (Es together, then subtract cases where Ds are together), part (b) requires calculating probability using specific positioning constraints, and part (c) involves selection with restrictions using complementary counting. While requiring organization and multiple techniques, these are standard A-level permutations methods without novel insight—slightly easier than average due to clear structure and routine application of formulas.
Spec5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems
  1. Find the number of different arrangements of the 9 letters in the word DELIVERED in which the three Es are together and the two Ds are not next to each other. [4]
  2. Find the probability that a randomly chosen arrangement of the 9 letters in the word DELIVERED has exactly 4 letters between the two Ds. [5]
Five letters are selected from the 9 letters in the word DELIVERED.
  1. Find the number of different selections if the 5 letters include at least one D and at least one E. [3]
Question 7:
AnswerMarks
7(a)Method 1: Arrangements with 3 Es together – arrangements with 3 Es together and 2 Ds together
7!
−6!
AnswerMarks Guidance
2!B1 7!
– e, e a positive integer (including 0).
2!
AnswerMarks
M1f – 6!, f > 6!
M17! 6!
− , a,c = 1, 2 and b,d = 1, 3.
a!b! c!d!
AnswerMarks
1800A1
Method 2: Identified scenarios ^ EEE ^ ^ ^
65
5!
AnswerMarks Guidance
2B1 5! × j, j a positive integer (j = 1 may be implied).
M1k! 65 k! k! 6P 76
 , 6C ,  2 or k! ,
m! 2 m! 2 m! 2 n
k a positive integer (k = 1 may be implied),
m = 1, 2 n = 1, 2, 3.
AnswerMarks
M1m( m−1 )
k! k a positive integer > 1, m = 10, 9, 8, 7, 6 and
n
n = 1, 2.
AnswerMarks
1800A1
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
7(b)First 2 marks: Method 1 – Number of arrangements with 2 Ds in one position with 4 letters in between – repeats allowed
7! × 4 × 2M1 7! × s, s = positive integer > 1.
M1t! × 4 × 2, t = 8, 7, 6.
Condone t! × 8.
First 2 marks: Method 2 – Picking 2Ds, arranging 4 letters from remaining letters between and then arranging terms
7P 4!2!
AnswerMarks Guidance
4M1 7P a!b! , 1 ⩽ a ⩽ 6 and b = 1, 2, 3.
4
AnswerMarks
M17P 4!2!, c = 3, 4, 5.
c
First 2 marks: Method 3 – Identified scenarios involving Es between Ds
D ^ ^ ^ ^ D E E E = 4C × 4! × 4! ×2! = 1152
4
D E ^ ^ ^ D E E ^ = 4C × 4! × 4! × 3 ×2! = 13824
3
D E E ^ ^ D E ^ ^ = 4C × 4! × 4! × 3 ×2! = 20736
2
D E E E ^ D ^ ^ ^ = 4C × 4! × 4! ×2! = 4608
AnswerMarks Guidance
1M1 1 identified scenario value correct.
M14 appropriate scenarios added, no incorrect.
QuestionAnswer Marks
7(b)Final 3 marks for Methods 1, 2 and 3
40320A1 If A0 scored, SC B1 for 40320 WWW.
 9!=
AnswerMarks Guidance
[Total number of arrangements =] 362880B1 Accept unsimplified.
May be seen as denominator of probability.
40320 1
Probability = =
AnswerMarks Guidance
362880 9B1FT their 40320
, accept unsimplified.
their 362880
B1FT if their 40320 and their 362880 supported by work in
this part.
Condone their 362880 supported by calculation in 7(a).
5
AnswerMarks
7(c)Scenarios
D E _ _ _ 4C 4
3
D E E _ _ 4C 6
2
D E E E _ 4C 4
1
D D E _ _ 4C 6
2
D D E E _ 4C 4
1
D D E E E [4C ] 1
AnswerMarks Guidance
0B1 1 correct unsimplified outcome/value for one identified
scenario excluding DDEEE.
Note: 4C cannot be used for 4C .
1 3
AnswerMarks
M1Add values of 6 appropriate scenarios, no additional, incorrect
or repeated scenarios. Accept unsimplified.
AnswerMarks
[Total =] 25A1
3
Question 7:
--- 7(a) ---
7(a) | Method 1: Arrangements with 3 Es together – arrangements with 3 Es together and 2 Ds together
7!
−6!
2! | B1 | 7!
– e, e a positive integer (including 0).
2!
M1 | f – 6!, f > 6!
M1 | 7! 6!
− , a,c = 1, 2 and b,d = 1, 3.
a!b! c!d!
1800 | A1
Method 2: Identified scenarios ^ EEE ^ ^ ^
65
5!
2 | B1 | 5! × j, j a positive integer (j = 1 may be implied).
M1 | k! 65 k! k! 6P 76
 , 6C ,  2 or k! ,
m! 2 m! 2 m! 2 n
k a positive integer (k = 1 may be implied),
m = 1, 2 n = 1, 2, 3.
M1 | m( m−1 )
k! k a positive integer > 1, m = 10, 9, 8, 7, 6 and
n
n = 1, 2.
1800 | A1
4
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | First 2 marks: Method 1 – Number of arrangements with 2 Ds in one position with 4 letters in between – repeats allowed
7! × 4 × 2 | M1 | 7! × s, s = positive integer > 1.
M1 | t! × 4 × 2, t = 8, 7, 6.
Condone t! × 8.
First 2 marks: Method 2 – Picking 2Ds, arranging 4 letters from remaining letters between and then arranging terms
7P 4!2!
4 | M1 | 7P a!b! , 1 ⩽ a ⩽ 6 and b = 1, 2, 3.
4
M1 | 7P 4!2!, c = 3, 4, 5.
c
First 2 marks: Method 3 – Identified scenarios involving Es between Ds
D ^ ^ ^ ^ D E E E = 4C × 4! × 4! ×2! = 1152
4
D E ^ ^ ^ D E E ^ = 4C × 4! × 4! × 3 ×2! = 13824
3
D E E ^ ^ D E ^ ^ = 4C × 4! × 4! × 3 ×2! = 20736
2
D E E E ^ D ^ ^ ^ = 4C × 4! × 4! ×2! = 4608
1 | M1 | 1 identified scenario value correct.
M1 | 4 appropriate scenarios added, no incorrect.
Question | Answer | Marks | Guidance
7(b) | Final 3 marks for Methods 1, 2 and 3
40320 | A1 | If A0 scored, SC B1 for 40320 WWW.
 9!=
[Total number of arrangements =] 362880 | B1 | Accept unsimplified.
May be seen as denominator of probability.
40320 1
Probability = =
362880 9 | B1FT | their 40320
, accept unsimplified.
their 362880
B1FT if their 40320 and their 362880 supported by work in
this part.
Condone their 362880 supported by calculation in 7(a).
5
--- 7(c) ---
7(c) | Scenarios
D E _ _ _ 4C 4
3
D E E _ _ 4C 6
2
D E E E _ 4C 4
1
D D E _ _ 4C 6
2
D D E E _ 4C 4
1
D D E E E [4C ] 1
0 | B1 | 1 correct unsimplified outcome/value for one identified
scenario excluding DDEEE.
Note: 4C cannot be used for 4C .
1 3
M1 | Add values of 6 appropriate scenarios, no additional, incorrect
or repeated scenarios. Accept unsimplified.
[Total =] 25 | A1
3
\begin{enumerate}[label=(\alph*)]
\item Find the number of different arrangements of the 9 letters in the word DELIVERED in which the three Es are together and the two Ds are not next to each other. [4]

\item Find the probability that a randomly chosen arrangement of the 9 letters in the word DELIVERED has exactly 4 letters between the two Ds. [5]
\end{enumerate}

Five letters are selected from the 9 letters in the word DELIVERED.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the number of different selections if the 5 letters include at least one D and at least one E. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE S1 2023 Q7 [12]}}
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