2014 A-level H2 Mathematics (9740) Paper 2 Question 6 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Required number of ways = {3 \choose 1} \times {8 \choose 4} \times {5 \choose 2} \times {6 \choose 4} = 31500

(ii)
Number of ways for both brother are in team = {3 \choose 1} \times {8 \choose 4} \times {4 \choose 1} \times {5 \choose 3} = 8400
Number of ways for both brother are not in team = {3 \choose 1} \times {8 \choose 4} \times {4 \choose 2} \times {5 \choose 4} = 6300
Required number of ways = 31500 - 8400 - 6300 = 16800 ways.

(iii)
We consider three cases here.
1. Remaining midfield plays as midfielder
2. particular midfield plays defender
3. particular midfield not playing

Required number of ways = {3 \choose 1} [{8 \choose 4} \times {3 \choose 1} + {8 \choose 3} \times {3 \choose 2} + {8 \choose 4} \times {3 \choose 2}] {5 \choose 4} = 8820 ways.

Personal Comments:
Students can solve (ii) by consider two cases, i.e., midfield bother is in team while attacker brother is not and midfield bother is not team while attacker brother i. (iii), I was a bit lazy to type it out to many times so I simply factorise everything, knowing that the number of ways to choose goalkeeper and attackers are the same.

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