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Mathematics 15 Online
OpenStudy (anonymous):

Show that \[12C5= 12C7\]

OpenStudy (anonymous):

Could that be a combination?

OpenStudy (anonymous):

yah sorry, i couldnt write it correctly

OpenStudy (anonymous):

\[nCr = \frac{ n! }{ r!(n-r)! }\]

ganeshie8 (ganeshie8):

yes, you may use the definition to prove. but to *see* they're equal, think of below : Number of ways of choosing 5 objects from 12 objects = Number of ways of leaving 7 objects behind

OpenStudy (anonymous):

Knowing that relation may help

OpenStudy (anonymous):

\[\frac{ 12! }{ 5!7! } = \frac{ 12! }{ 7!5! }\]

OpenStudy (anonymous):

@ganeshie8 is that correct?

OpenStudy (anonymous):

@Johnbc

OpenStudy (anonymous):

Correct

OpenStudy (anonymous):

You will now notice that you have the same equation on both sides so you have proven they are equal to each other

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