The diagram shows quadrilateral ABCD and the bisectors of the angles at A, B, C and D - HSC - SSCE Mathematics Extension 1 - Question 14 - 2018 - Paper 1

To find the coefficients of $y^{r}$ in the expansions of $(1 + (x + y)^{3})$ and $(2 + y)^{3}$, let's start with the binomial expansions:

1. For $(1 + (x + y)^{3})$:

   • Using the binomial theorem:
   • We expand $(x + y)^{3}$: $\binom{3}{0}x^{3}y^{0} + \binom{3}{1}x^{2}y + \binom{3}{2}xy^{2} + \binom{3}{3}y^{3}$
   • Then we have: $1 + \binom{3}{1}(x + y) + \binom{3}{2} \frac{(x + y)^{2}}{2!} + ...$

2. For $(2 + y)^{3}$:

   • Again applying the binomial theorem:
   • We have: $\binom{3}{0}2^{3} + \binom{3}{1}2^{2}y + \binom{3}{2}2y^{2} + \binom{3}{3}y^{3}$

Equating coefficients allows us to analyze how to find the corresponding terms in both expansions, particularly for specific values of $r$.

The selection process involves two selectors:

1. Selector A chooses a group of at least 4 people from the 23 applicants:

   • The number of ways in which Selector A can choose a group of n people from 23 is given by: $\binom{23}{n}$ where $n \geq 4$.

2. Selector B then chooses 4 people from the selected group by Selector A:

   • For each chosen group of size n, the number of ways to select 4 from n is given by: $\binom{n}{4}$

To find the total ways this selection can be made, we sum over all valid n:

$\text{Total ways} = \sum_{n=4}^{23} \binom{23}{n} \cdot \binom{n}{4}$

Calculating these terms can be computed using combinatorial identities to arrive at the final answer.