1.
3
6
3   3x 
Term in x  C 3  2 

 2 
Note:
3
(M1)(A1)(A1)(A1)
Award M1 for recognizing Binomial
Theorem and A1 for each correct
element.
 27 x 3 

8 

= 20  8   
A1
=  540 x 3
A1
(Coefficient of x = 540)
3
2.

  3   3 3   2  3 3  2
3
3
3 2 
2
1
[6]
2
  2
3
M1A1
= 3 3   18  12 3  8
= 15 3  26
(M1)A1
(accept a = 15, b = 26)
A1A1
N3
[6]
3.
(a)
(b)
5
5
4
3 2
2 3
4
5
(2 + x) = 2 + 5(2) (x) + 10(2) x + 10(2) x + 5(2)x + x
(M1)(A1)
= 32 + 80x + 80x2 + 40x3 + 10x4 + x5
(A1) (C3)
Note: Award (C2) for 5 correct terms, (C1) for 4 correct terms.
Let x = 0.01 = 10–2
 (2.01)5 = 32 + 0.8 + 0.008 + 0.000 04 + 0.000 0001 + 0.000 000 0001(M1)(A1)
= 32.808 040 1001
(A1) (C3)
[6]
4.
 8  – 1 
3
The coefficient of x3 is  

 3  2 
(M2) (A2)
The coefficient of x3 is –7
(A2) (C6)
[6]
5.
The total number of four-digit numbers = 9 × 10 × 10 × 10 = 9000.
The number of four-digit numbers which do not contain a digit 3
= 8 × 9 × 9 × 9 = 5832.
Thus, the number of four-digit numbers which contain at least one digit 3 is
9000 – 5832 = 3168.
(A1)
(A1)
(A1) (C3)
[3]
6.
METHOD 1
Consider the group as two groups – one group of the two oldest
and one group of the rest
Either one of the two oldest is chosen or neither is chosen
Then the number of ways to choose the committee is
 6  2   6  2 
      
 3  1   4  0 
= 40 + 15
= 55 ways
METHOD 2
The number of ways to choose a committee of 4 minus the number
of ways to have both the oldest 
8 6
    
 4  2
= 70 – 15
= 55 ways
(M2)
(M1)
(M1)(M1)
(A1) (C6)
(M3)
(M1)(M1)
(A1) (C6)
[6]
1
7.
9
The first car can be filled in
C 3 ways.
The second car can be filled in
The third car can be filled in
Number of combinations =
9
3
6
M1A1
C 3 ways.
A1
C 3 ways.
(A1)
C 3  C 3  C 3 = 84  20  1 = 1680.
6
3
M1A1
N4
[6]
2