Australian Mathematics Competition 2021 – Intermediate, Problem 26

Let's discuss a problem from the AMC 2021 Intermediate: Problem 26 which revolves around basic algebra and combinatorics.

Problem

In Australian Rules football, a team scores six points for a 'goal' and one point for a 'behind'. During a game, Vladislav likes to record his team's score with a sequence of sixes and ones. There are exactly three distinct sequences which give a final score of 7 points, namely 6,1 and 1,6 and (1,1,1,1,1,1,1). How many different sequences provide a final score of 20 points?

Solution

Let x be the no. of sixes and y be the no. of ones. Thus the equation for 20 points becomes:

\( 6x + y = 20\)

As \(y\) is the number of behinds, we know \(y \geq 0\), and \(x\) must be a non-negative integer such that \(6 x+y=20\). To find all possible values of \(x\), we can solve for \(y\) :

\(y = 20 - 6x\)

Since , \( 6x \leq 20\), so the possible values of \(x\) be \(0, 1, 2\), and \(3\).

For, \(x = 0\), \(y = 20\),

For, \(x = 1\), \(y = 20 - 6 = 14\),

For, \(x = 2\), \(y = 20 - 12 = 8\),

For, \(x = 3\), \(y = 20 - 18 = 2\).

Now for each pair of \(x,y\) the number of distinct sequences is the number of ways to arrange x sixes and y ones. This is a combinatorial problem where the binomial coefficient gives the number of distinct sequences,

\({x + y}\choose{x}\) = \(\frac{(x+ y)!}{x!.y!}\)

For \(0,20\) ; \({20}\choose{0}\) = 1,

For \(1,14\) ; \({15}\choose{1}\) = 15,

For \(2,8\) ; \({10}\choose{2}\) = 45,

For \(3,2\) ; \({5}\choose{3}\) = 10.

Thus the sum of the number of sequences = \(1 + 15 + 45 + 10 = 71\).

What is AMC (Australian Mathematics Competition)?

The Australian Mathematics Competition (AMC) is one of Australia's largest and oldest annual mathematics competitions, aimed at fostering interest and excellence in mathematics among students.