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What is the value of n(C)r when n=8 and r=7?
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I believe it would be 8. 8(C)7 = (8!)/(7! x 1!) = 8
When you get to the (7! x1!) you just factor out the 7 correct?
I would simplify the denominator to 7! as 1! is just 1. 8! = 8 x 7 x 6 ...1 7! = 7 x 6 x 5 ... 1 Therefore everything would cancel except for the eight in the numerator.
Alright, that's what I had just wasn't sure. Thank you!
\[nCr=\frac{nPr}{r!}\]
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i remember that easier than the other method; 8P7 is just start at 8 and pick 7 8.7.6.5.4.3.2 diivide by 7! 8.7.6.5.4.3.2 ----------- = 8 7.6.5.4.3.2
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