Difference between revisions of "2020 AMC 10B Problems/Problem 19"
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<math>\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}</math> | <math>\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}</math> | ||
− | We need to get rid of the multiples of <math>3</math>, which will subsequently get rid of the multiples of <math>9</math> | + | We need to get rid of the multiples of <math>3</math>, which will subsequently get rid of the multiples of <math>9</math> (if we didn't, the zeroes would mess with the equation since you can't divide by 0) |
<math>9\cdot5=45</math>, <math>8\cdot6=48</math>, <math>\frac{51}{3}</math> leaves us with 17. | <math>9\cdot5=45</math>, <math>8\cdot6=48</math>, <math>\frac{51}{3}</math> leaves us with 17. |
Revision as of 16:19, 7 February 2020
Solution
We're looking for the amount of ways we can get cards from a deck of
, which is represented by
.
We need to get rid of the multiples of , which will subsequently get rid of the multiples of
(if we didn't, the zeroes would mess with the equation since you can't divide by 0)
,
,
leaves us with 17.
Converting these into, we have
~quacker88