Combinatorics — Chain for n=4

Question

(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).

This practice question was inspired by Question 14(a) in the 2020 NSW HSC Mathematics Extension 1 examination, © NESA. It is an original TestMum variation that practises the same mathematical idea, not an official NESA question.

Step by step

Set up the solution: count a simple unrestricted set first, then add or subtract the arrangements that break the condition.(i) Coefficient of on the left is .
Simplify the previous line carefully, keeping exact values where possible.On the right: .
Simplify the previous line carefully, keeping exact values where possible.(ii) For each , choose men from and women from in ways; sum and apply (i) to get .
Simplify the previous line carefully, keeping exact values where possible.(iii) After choosing of each gender (), choose leaders in ways. Sum from to .
Finish the calculation, then check that the result meets the question’s conditions.(iv) Leaders first: choices for the man and for the woman. Remaining: balanced groups from of each gender, which by (ii) with is ways.
State the final answer clearly in the form the question requested.Therefore .

Common mistake: changing so many features that the new example no longer tests the same mathematical idea.

Verification: Direct: .

Any mark labels are a TestMum study aid, not an official NESA marking allocation.

← Back to 2020 HSC Extension 1 paper Question 14(a)