Question
(a) Find the number of different arrangements of the 9 letters in the word ANDROMEDA in which no consonant is next to another consonant. (The letters D, M, N and R are consonants and the letters A, E and O are not consonants.) [3]
(b) Find the number of different arrangements of the 9 letters in the word ANDROMEDA in which there is an A at each end and the Ds are not together. [3]
Four letters are selected at random from the 9 letters in the word ANDROMEDA.
(c) Find the probability that this selection contains at least one D and exactly one A. [4]
Additional Page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a) 5! ×4! M1 5! × 4!
, e a positive integer, 1 can be
2! × 2! e
implied. No other terms on numerator. No
addition etc.
M1 f
, f a positive integer, g = 1, 2. No
2! × g !
other terms on denominator.
720 A1
3
7(b) Method 1 Number of arrangements with A at each end – Number of arrangements with A at each end and 2 Ds together.
7! B1 7!
− 6! – e, 7 P5 − e, e a positive integer or 0.
2! 2!
M1 6!
d– , d > 720, r = 1, 2.
r!
= 1800 A1
Method 2 A ^ ^ ^ ^ ^ A and Ds inserted separately
6
P2 6×5 B1 5! × s, s a positive integer, 1 may be implied.
5! × or 5! × or 5! × 6C2
2! 2
M1 6×5
t× , t a positive integer > 1, u =1, 2.
u
= 1800 A1
Method 3 Number of arrangements with As at each end and Ds placed in different scenarios.
Scenario position of first D B1 Correct outcome/value for 1 identified
scenario, accept unsimplified, www.
AD^^^^^^A 5! × 5 600
A^D^^^^^A 5! × 4 480 M1 Add values of 5 correct scenarios, no
incorrect/repeated scenarios.
A^^D^^^^A 5! × 3 360
A^^^D^^^A 5! × 2 240
A^^^^D^^A 5! × 1 120
[Total =] 1800 A1
3
7(c) Method 1:
M1 At least one correct unsimplified expression
Scenarios for an identified scenario.
AD^^ 2
C1 × 2C1 × 5C2 = 40
ADD^ 2
C1 × [ 2C2 ×] 5C1 = 10
[Total = ] 40 + 10 or 50 soi A1 www
If M0 scored, SC B1 [total =]50 www.
[Total number of selections =] 9C4 [= 126] B1 Accept evaluated, accept as denominator of
probability expression.
Do not condone 9C5 unless there is a clear
explanation for selecting the letters not in the
group.
50 25 B1 FT 0.396825… to at least 3SF.
[Probability =] ,
126 63
their attempted 40 + 10
FT . Numerator must
126
be from an attempt to find the 2 appropriate
scenarios and must be evaluated.
7(c) Method 2:
M1 Numerator for at least one correct
Scenarios
unsimplified expression for an identified
AD^^ scenario.
2 2 5 4 4 960 20
× × × × P2 = , 2 × 2 × 5 × 4 × 12 2 × 2 ×1× 5 ×12
9 8 7 6 3024 63 either or
a×b×c×d a×b×c×d
ADD^ seen, 6 ⩽ a,b,c,d ⩽ 9.
2 2 1 5 4 P3 240 5
× × × × = ,
9 8 7 6 2! 3024 63 A1 2 × 2 × 5 × 4 ×12 2 × 2 ×1× 5 ×12
+ ,
a×b×c×d a×b×c×d
20 5 6 ⩽ a,b,c,d ⩽ 9.
[Total Probability = ] + 1200 25
63 63 If M0 scored, SC B1 , g > 1200, or
g 63
seen.
B1 p q r s
× × × present in all scenarios
9 8 7 6
t
attempted, accept , t < 3024.
3024
1200 25 B1 FT 0.396825… to at least 3SF.
, oe
3024 63
their attempted 960 + 240
FT . Numerator
3024
must be from an attempt to find the 2
appropriate scenarios.
7(c) Method 3: selecting the A and then selecting 3 any letters and removing selections without Ds.
2
C1 × (7C3 – 5C3) [= 2 × (35 – 10)] M1 a × (7C3 – 5C3), a = 1, 2.
[Total = ] 50 A1 www
If M0 scored, SC B1 [total =]50 www.
[Total number of selections =] 9C4 [= 126] B1 Accept evaluated, accept as denominator of
probability expression.
Do not condone 9C5 unless there is a clear
explanation for selecting the letters not in the
group.
50 25 B1 FT 0.396825… to at least 3SF.
[Probability =] ,
126 63 their attempted 40 + 10
FT . Numerator must
126
be from an attempt to find the 2 appropriate
scenarios.
Method 4: Listing outcomes.
Either 10 correct outcomes for ADD^ listed or 40 correct outcomes for AD^^ listed M1
50 stated A1 www
If M0 scored, SC B1 [total =]50 www.
126 stated or correct outcomes listed B1
50 25 B1 0.396825… to at least 3SF.
[Probability =] ,
126 63 their attempted 40 + 10
FT . Numerator must
126
be from an attempt to find the 2 appropriate
scenarios.
4