Find the number of selections Grace could make if the largest integer drawn among the four cards is either a \(5\), a \(6\) or a \(7\).

Find the number of selections Grace could make if at least two of the four integers drawn are even.

use of the addition principle with \(3\) terms     (M1)

to obtain \(^4{C_3}{ + ^5}{C_3}{ + ^6}{C_3}{\text{ }}( = 4 + 10 + 20)\)     A1

number of possible selections is \(34\)     A1

[3 marks]

EITHER

recognition of three cases: (\(2\) odd and \(2\) even or \(1\) odd and \(3\) even or \(0\) odd and \(4\) even)     (M1)

\(\left( {^5{C_2}{ \times ^4}{C_2}} \right) + \left( {^5{C_1}{ \times ^4}{C_3}} \right) + \left( {^5{C_0}{ \times ^4}{C_4}} \right)\;\;\;( = 60 + 20 + 1)\)     (M1)A1

OR

recognition to subtract the sum of \(4\) odd and \(3\) odd and \(1\) even from the total     (M1)

\(^9{C_4}{ - ^5}{C_4} - \left( {^5{C_3}{ \times ^4}{C_1}} \right)\;\;\;( = 126 - 5 - 40)\)     (M1)A1

THEN

number of possible selections is \(81\)     A1

[4 marks]

Total [7 marks]

As the last question on section A, candidates had to think about the strategy for finding the answers to these two parts. Candidates often had a mark-worthy approach, in terms of considering separate cases, but couldn’t implement it correctly.

As the last question on section A, candidates had to think about the strategy for finding the answers to these two parts. Candidates often had a mark-worthy approach, in terms of considering separate cases, but couldn’t implement it correctly.