Question

A hockey game between two teams is 'relatively close' if the number of goals scored by the two teams never differ by more than two. In how many ways can the first 12 goals of a game be scored if the game is 'relatively close'?

Answer 1

I assume that the game is relatively close at the 12th goal mark, so we can rule out something like 12 goals by one team followed up by 12 goals by the other team.

Answer 2

i can give u the answer can u figure out the working out for me?

Answer 3

We have 6-6, 5-7, and 7-5 as possible scores.

Answer 4

The number of ways these scores can appear is basically designated by combinations.
Our answer is basically \[
{12 \choose 7} + {12\choose 6} + {12\choose 5}
\]

Answer 5

sorry but this is wrong, answer is over 900

Answer 6

My answer is over 2000. What is the problem?

Answer 7

this answer is below 1000, over 900

Answer 8

if they score evenly, then it is just \[
{12 \choose 6} = 924
\]

Answer 9

answer is 972

Answer 10

can u try to figure out the working out

Answer 11

how do they define a "way"?

Answer 12

i got the answer but i dont know the working out

Answer 13

Give some examples of a 'way' in which teams can score.

Answer 14

1-0, 1-1, 2-1, 2-2, 3-2 until the total score gets to 12, i think....

Answer 15

That can't be right. Even if you consider the score n-m, where n is 0 to 12 and m is 0 to 12, you would get 13 * 13 = 169 ways.

Answer 16

Consider if there is team 1 and team 2. If you have only 4 games... we could right the score history as follows:
1111
1112
1121
...
2222

Answer 17

keep going, sir

Answer 18

For 12 games, we can see it as 12 boxes that need to have a 1 or 2.
_ _ _ _ _ _ _ _ _ _ _ _
Suppose we fill in 6 of the boxes:
_ _ _ 1 1 1 _ _ 1 _ 1 1

There is only one way to fill in the remaining boxes:
2 2 2 1 1 1 2 2 1 2 1 1

Answer 19

Deciding which box to fill with team 1 is just a matter of choose 6 boxes out of 12, not caring about the order of the boxes. This is combinations. It is represented by \(^{12}C_6\).

Answer 20

by the way, i cant use a calculator for this question

Answer 21

972 doesn't make any sense to me.

Answer 22

well this is australian mathematics competition

Answer 23

@wio never differ by more than two at any point in time, even while the game is ongoing

Answer 24

basically the problem is asking to evaluate 2(a+b) where a->2(a+b) and b->a+b 5 times which is just 2*3^5*(a+b), which evaluated at a=1, b=1 is 972

2(a+b) represents the first two points of the game, each successive two points are taken care by a->2(a+b) and b->a+b representing
A. a state in which one of the players has a 2-point advantage or
B. a state in which both players are again at a draw

note that it is impossible for a player to have a one-point advantage, as expected in a 12-set game

Answer 25

Okay, this problem is kinda lame, but I got an answer.

Answer 26

*my previous post should say a->a+b and b->2(a+b)

Answer 27

You have to consider that team 1 has some score, and the other team has -2, -1, 0, +1, or +2 relatively speaking.

So you have this table: \[
\begin{array}{r|c|c|c|c|c}
scores&-2&-1&0&+1&+2 \\
\hline
0&0&0&1&0&0\\
1&0&1&0&1&0\\
2&1&0&2&0&1\\
3&0&3&0&3&0\\
4&3&0&6&0&3\\
5&0&9&0&9&0\\
6&9&0&18&0&9\\
7&0&27&0&27&0\\
8&27&0&54&0&27\\
9&0&81&0&81&0\\
10&81&0&162&0&81\\
11&0&243&0&244&0\\
12&243&0&486&0&243
\end{array}
\]Then add up \(243+486+243 = 972\)

Answer 28

you don't need to fill out the table since the sum of row n=3*the sum of row n-2, this is basically a variant of pascal's triangle & similar identities apply

Answer 29

after adding up the numbers in the scores=2 row, multiply by 3^5 to get the sum for scores=12

Answer 30

What is a and b?

Answer 31

a and b are just dummy variables representing states at which
A)one team is at a 2-point advantage over the other team and
B)both teams have an equal number of points
that's why they are set to 1 at the end

Answer 32

if both teams have an equal number of points (state B), the next two points can play out in the following ways: 1)team #1 gets both points, leading to state B; 2)team #2 gets both points, leading to state B; 3)team #1 gets the first point, team#2 gets the second, leading to state A; 4)team #2 gets the first point, team #1 gets the second, leading to state A
mathematically, B->2(A+B)

if one team has a 2-point advantage (state A), the next two points can play out in the following ways: 1)the team with the advantages loses both points, leading to state B; 2)the team with the advantage loses only the first point, leading to state A again
mathematically, A->A+B

Answer 33

we start in state B (at a tie). B->2(A+B) after the first two points
2(A+B) -> 6(A+B) after the next two points
6(A+B) -> 18(A+B) after the next two points
and so forth up to 972(a+b)=972 for the 12th point