Difference between revisions of "2019 AMC 10C Problems/Problem 11"

Revision as of 19:48, 27 January 2020

We could use casework:

1 1 1 1 1 1 1 1 1 1 = 10 choose 0 = 1 case

1 1 1 1 1 1 1 1 2 = 9 choose 1 = 9 cases

1 1 1 1 1 1 2 2 = 8 choose 2 = 28 cases

1 1 1 1 2 2 2 = 7 choose 3 = 35 cases

1 1 2 2 2 2 = 6 choose 2 = 15 cases

2 2 2 2 2 = 5 choose 0 = 1 case

We add up 1 + 9 + 28 + 35 + 15 + 1 = $89$ cases.

~rg_ryse

Revision as of 19:46, 27 January 2020 (view source) Rg ryse (talk \| contribs) (Solution)		Revision as of 19:48, 27 January 2020 (view source) Rg ryse (talk \| contribs) (→‎Solution) Newer edit →
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	We add up 1 + 9 + 28 + 35 + 15 + 1 = <math>89</math> cases.		We add up 1 + 9 + 28 + 35 + 15 + 1 = <math>89</math> cases.
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		+	~rg_ryse