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Title: Consider the following formulas: nPr = (n!/(n-r)!) and nCr = (n!/(n-r)!r!). Giv Post by: ryobu on Sep 27, 2015 Consider the following formulas: nPr = (n!/(n-r)!) and nCr = (n!/(n-r)!r!).
Given the same values for n and r in each formula, which is the smaller value, P or C? How does this relate to the concept of counting the number of outcomes based on whether or not order is a criterion? Title: Re: Consider the following formulas: nPr = (n!/(n-r)!) and nCr = (n!/(n-r)!r!). Giv Post by: padre on Sep 29, 2015 The combination value will be smaller, since order is not important. For example, ABC is equivalent to ACB and would not be counted twice. If, however, r is 0 or 1 then nPr = nCr.
Title: Re: Consider the following formulas: nPr = (n!/(n-r)!) and nCr = (n!/(n-r)!r!). Giv Post by: ryobu on Oct 10, 2015 If only I had more time to submit my work, I would have gotten it right :-\ Your answer was right.
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