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'?

Question

anonymous · Answer

@BangkokGarrett  @CGGURUMANJUNATH @carebr0022 @djoplin @eSpeX @Fifciol @genius12 @ganeshie8

anonymous · Answer

@Hero

anonymous · Answer

Anyone got a answer apart from @ChrisJP? ^
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hero · Answer

I get 30 total possible ways....

1 0 
0 1
0 2
2 0
1 1
1 2
2 1
1 3
3 1 
2 2
2 3
3 2
2 4
4 2
3 3
3 4
4 3
4 4
4 5
5 4
3 5
5 3
5 5
4 6
6 4
6 5
5 6
6 6
7 5
5 7

anonymous · Answer

I tried to take on this one by realising the scores have to be either 5 and 7, 6 and 6 or 7 and 5. I then did (2^5+2^7)+(2^6+2^6)+(2^7+2^5) and got 448 (this was the only one i could attempt). Is this correct?

hero · Answer

@philliphoang did you have anything to add?

anonymous · Answer

Well you had a answer of 30 but @opensessame had a answer of 448 So Um im not sure who got it right:\

hero · Answer

I'm pretty sure mine is reasonable enough. It's pretty self explanatory. There are only two teams and obviously "relatively close" applies to more than just 

5 7
6 6
7 5

They wanted all possible ways of the first 12 goals being 'relatively close'

Not scores that add to 12 only. Furthermore, there doesn't seem to be any reasonable explanation for @OpenSessame result.

hero · Answer

The term "relatively close" was clearly defined in the problem.

anonymous · Answer

how many ways can the first 12 goals of a game be scored if the game is 'relatively close'?
That means that every score could be different...

hero · Answer

@OpenSessame, did you understand the solution I posted at all?