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Show that \[12C5= 12C7\]
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Could that be a combination?
yah sorry, i couldnt write it correctly
\[nCr = \frac{ n! }{ r!(n-r)! }\]
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
Knowing that relation may help
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\[\frac{ 12! }{ 5!7! } = \frac{ 12! }{ 7!5! }\]
@ganeshie8 is that correct?
@Johnbc
Correct
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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