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use combinations

@pgpilot326 u mean

|dw:1444502981895:dw|

i know that the fomular for combination is : \[\frac{ n! }{ r!(n-1) }\]

|dw:1444503437050:dw|

btw\[\left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{ n! }{ r!\,\,\left( n-r \right)! }\]

there like 13 red ball and 17 yellow

okay and how many of each are needed?

13 red we 3 and 17 yellow for 1

3 red? i don't think that's right

misread is 2 red

correct. so how many ways to select 2 of the 13?

26

no, use combinations. you are sampling without replacement and order doesn't matter

ok

is 286 |dw:1444503856355:dw|

yep and how many ways to select 1 of the yellow balls?

17

yep but use combinations

|dw:1444503992842:dw|

|dw:1444504035212:dw|

now we have like 286 for red and 17 for yellow

\[\frac{ 4862 }{ 9880 }\]