2020 AMC 10B Problems/Problem 19: Difference between revisions
| Line 7: | Line 7: | ||
<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 15:19, 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
