Make £20.17

First let us consider the use of 1p coins. The number of 1p coins used must be $n_{1p} = 2017-5k$ , where $k=0,1,2,...403$ and $n_{1p} = 2,7,12,...2017$ , a total of 403 ways. For each way or each $k$ , there is a combination of 5p and 10p coins used or $(n_{5p}, n_{10p})$ . For example, when $k=0$ , the only combination possible is $(n_{1p},n_{5p}, n_{10p}) = (2017,0,0)$ and the number of way is $N_0 = 1$ . We note that $N_k$ is the same as the number of ways to use 5p ( $N_{5,k}$ ) and that of 10p ( $N_{10,k}$ ). For $k=0$ , $N_{5,0}=N_{10,0} = 1$ because there is one way of using 5p (use zero 5p) and using 10p (use zero 10p). The combinations and $N_k$ and $N_{10,k}$ of the first few $k$ are given below to illustrate this point:

\(\begin{array} {} k = 0 & (2017,0,0) & N_0 = 1 & N_{10,0} = 1 \\ k = 1 & (2012,1,0) & N_1 = 1 & N_{10,1} = 1 \\ k = 2 & (2007,2,0), (2007,0,1) & N_0 = 2 & N_{10,2} = 2 \\ k = 3 & (2002,3,0), (2002,1,1) & N_3 = 2 & N_{10,3} = 2 \\ k = 4 & (1997,4,0), (1997,2,1), (1997,0,2) & N_4 = 3 & N_{10,4} = 3 \end{array} \)

Now, for a $k$ , The number of ways to use 10p $N_{10,k} = N_k = \left \lfloor \dfrac k2 \right \rfloor + 1$ , where $\lfloor \cdot \rfloor$ denotes the floor function . Therefore, the total number of ways:

$\begin{aligned} N & = \sum_{k=0}^{403} \left( \left \lfloor \dfrac k2 \right \rfloor + 1 \right) = 2\sum_{k=1}^{201} k + 404 = 201 \times 202 + 404 = \boxed{41006} \end{aligned}$