2020 HSC Mathematics Extension 1
Question 14(a):
Combinatorics
(i) Use the identity to show that , where is a positive integer.
(ii) A club has members, with women and men. A group consisting of an even number () of members is chosen, with the number of men equal to the number of women. Show that the number of ways to do this is .
(iii) From the group chosen in part (ii), one of the men and one of the women are selected as leaders. Show that the number of ways to choose the even number of people and then the leaders is .
(iv) The process is now reversed so that the leaders, one man and one woman, are chosen first. The rest of the group is then selected, still made up of an equal number of women and men. By considering this reversed process and using part (ii), find a simple expression for the sum in part (iii).
How to recognise this question
Question type: combinatorics
- The command and notation point to combinatorics.
- The word “hence” means the later part is intended to reuse the earlier result.
How to handle it: count a simple unrestricted set first, then add or subtract the arrangements that break the condition.
Watch out: Do not restart the second part with unrelated numbers or discard the result proved in the first part.
Step-by-step answer
- (i) Coefficient of on the left of is .
- On the right, arises from products , and , giving .
- NESA (i): 2 marks; 1 for identifying the left-hand coefficient.
- (ii) For each , choose men from and women from in ways. Sum and apply (i) to get .
- NESA (ii): 2 marks.
- (iii) After choosing men and women, choose a male leader in ways and a female leader in ways, giving weight . Sum from to .
- NESA (iii): 2 marks.
- (iv) Choose the two leaders first: choices for the man and for the woman. Remaining: choose an equal number from the leftover of each gender, which by (ii) can be done in ways.
- Therefore .
- NESA (iv): 2 marks; 1 for recognising the reduction to part (ii) with .
Common mistake: choosing a familiar formula before checking that its conditions, signs, interval and units match this question.
The paper records 8 total marks for this group. Where the official marking guide provides useful partial-credit criteria, those criteria are stated in the steps above.
Try your hand at it again by answering these questions
Question 1
(i) Use the identity to show that .
(ii) A club has members, with women and men. A group consisting of an even number of members is chosen, with equally many men and women. Show that the number of ways to do this is .
(iii) From the group chosen in part (ii), one of the men and one of the women are selected as leaders. Show that the number of ways to choose the even group and then the leaders is .
(iv) The process is now reversed: leaders (one man and one woman) are chosen first, then the rest of the balanced group. Using part (ii), find a simple expression for the sum in part (iii).
Question 2
(i) Use the identity to show that .
(ii) A club has members, with women and men. A group consisting of an even number of members is chosen, with equally many men and women. Show that the number of ways to do this is .
(iii) From the group chosen in part (ii), one of the men and one of the women are selected as leaders. Show that the number of ways to choose the even group and then the leaders is .
(iv) The process is now reversed: leaders are chosen first, then the rest of the balanced group. Using part (ii), find a simple expression for the sum in part (iii).
Question 3
(i) Use the identity to show that .
(ii) A club has members, with women and men. A group consisting of an even number of members is chosen, with equally many men and women. Show that the number of ways to do this is .
(iii) From the group chosen in part (ii), one of the men and one of the women are selected as leaders. Show that the number of ways to choose the even group and then the leaders is .
(iv) The process is now reversed: leaders are chosen first, then the rest of the balanced group. Using part (ii), find a simple expression for the sum in part (iii).