# Problem of the Week Problem D and Solution The Dart Game

## Problem

A carnival dart game has three non-overlapping circles in a rectangle. One circle has a value of $$2$$, another has a value of $$3$$, and the third has a value of $$5$$. You are allowed to throw up to $$10$$ darts, and you start the game with a running total of 0. If a dart lands in one of the circles, you add the value of the circle to the running total. If a dart does not land in one of the circles, then you do not add anything to the running total for that throw.

Suppose you have exactly $$30$$ points after $$10$$ throws. Let $$a$$ represent the number of throws that landed in the circle with value $$5$$, let $$b$$ represent the number of throws that landed in the circle with value $$3$$, and let $$c$$ represent the number of throws that landed in the circle with value $$2$$. Determine all possibilities for $$(a,b,c)$$.

## Solution

We need to determine all possibilities for $$(a,b,c)$$ with $$5a+3b+2c=30$$ and $$a+b+c\leq 10$$.

We will look at cases for $$a$$. Since $$6 \times 5 = 30$$, then the largest value of $$a$$ is $$6$$. The smallest value is $$a=0$$ since $$a\geq 0$$.

Letâ€™s look at the specific case where $$a=2$$ to develop a process for how to determine the number of ways to get a total of $$30$$. We will use this process for all cases, but will not show our steps in the other cases.

If $$a=2$$, this will account for a total of $$2\times 5=10$$ points. Therefore, $$30-10=20$$ points will be needed from landing in the circles with values of $$2$$ and $$3$$.

Next, we find the maximum value for $$b$$, the number of throws that landed in the circle with value $$3$$. We want $$b$$ to account for a total that is less than or equal to $$20$$, but also give a remainder that is even, since the remaining points need to come from the circle with value $$2$$.

If $$b=7$$, then this would give $$7\times 3 = 21$$ points, which exceeds $$20$$. When $$b=6$$, then this would give $$6 \times 3 = 18$$ points. Then $$c=1$$ would make the total exactly $$30$$. Notice here that $$a+b+c = 2 + 6 + 1 = 9 \leq 10$$, as required. Thus, one possibility is that $$a=2$$, $$b= 6$$, and $$c=1$$.

We then need to replace circles with a value of $$3$$ with circles with a value $$2$$. We note that for every two circles with a value of $$3$$, we have a total value of $$6$$. We can replace those two circles with three circles of value $$2$$. This means that $$b=6-2 = 4$$ and $$c = 1 + 3 = 4$$. Notice here that $$a+b+c = 2 + 4 + 4 = 10 \leq 10$$, as required. Thus, another possibility is that $$a=2$$, $$b= 4$$, and $$c=4$$.

We can again replace two circles with a value of $$3$$ with three circles of value $$2$$. This means that $$b=4-2 = 2$$ and $$c = 4 + 3 = 7$$. Notice here that $$a+b+c = 2 + 2 + 7 = 11 >10$$. Therefore, this is not a possibility.

We can again replace two circles with a value of $$3$$ with three circles of value $$2$$. This means that $$b=2-2 = 0$$ and $$c = 7 + 3 = 10$$. Notice here that $$a+b+c = 2 + 0 + 10 = 12 >10$$. Therefore, this is not a possibility.

We cannot again replace two circles with a value of $$3$$ with three circles of value $$2$$ since this would make $$b$$ negative.

To summarize, when $$a=2$$, there are two combinations that give a total of $$30$$ and have $$a+b+c \leq 10$$.

We use this process for all the possible values of $$a$$. Our results are summarized in the table below.

$$a$$ $$b$$ $$c$$ $$5a+3b+2c$$ $$a+b+c$$
$$6$$ $$0$$ $$0$$ $$30$$ $$6$$
$$5$$ $$1$$ $$1$$ $$30$$ $$7$$
$$4$$ $$2$$ $$2$$ $$30$$ $$8$$
$$4$$ $$0$$ $$5$$ $$30$$ $$9$$
$$3$$ $$5$$ $$0$$ $$30$$ $$8$$
$$3$$ $$3$$ $$3$$ $$30$$ $$9$$
$$3$$ $$1$$ $$6$$ $$30$$ $$10$$
$$2$$ $$6$$ $$1$$ $$30$$ $$9$$
$$2$$ $$4$$ $$4$$ $$30$$ $$10$$
$$1$$ $$7$$ $$2$$ $$30$$ $$10$$
$$0$$ $$10$$ $$0$$ $$30$$ $$10$$

We find that there are $$11$$ possibilities for the number of throws that have landed in each circle. The $$11$$ possibilities for $$(a, b,c)$$ are:

$$(6,0,0)$$, $$(5, 1, 1)$$, $$(4, 2, 2)$$, $$(4, 0, 5)$$, $$(3, 5, 0)$$, $$(3, 3, 3)$$, $$(3, 1, 6)$$, $$(2, 6, 1)$$, $$(2, 4, 4)$$, $$(1, 7,2)$$, $$(0,10,0)$$