Date: Tue, 15 Nov 1994 10:16:54 -0500 From: steve barkin Subject: Chris Grant's factorial issue Dear Dr. Math My name is Chris Grant. I am using my teacher's mailbox with his permission. One day when I was playing on a calculator, I found that 1 and 0 both have a factorial of 1. Can you explain this? Thank you, Chris Grant. Steve Barkin Graham & Parks School Cambridge, Ma 02139
Date: Mon, 15 Nov 1993 16:30:29 -0500 From: Demetri Bonaros Subject: Re: Chris Grants factorial issue Hello, Chris! Thank you for writing to Dr. Math. Your question is indeed a very good one, one that puzzles a lot of students when they are first introduced to the fact that 0!=1. I think there a couple of reasons that explain this fact; offhand I can only think of one of them, but I'm pretty sure that it's not the only one. First, let's see what the binomial coefficients are. We define them as follows (OK, this is going to be a little hard to do on email, but here goes, anyway) /k\ \i/ (the / and\ are supposed to form one big parenthesis) is called "k choose i". Its value is the answer to the question "in how many ways can we choose i objects out of a set of k?" The value of "k choose i" is: k! __________ (can you prove this? Try induction) i!(k-i)! Let's see a few examples: "k choose 1" = in how many ways can you choose 1 object from a set of k? Well, the answer has to be k, right? If you have k objects, you have k choices. "k choose (k-1)" = in how many ways can you choose (k-1) objects out of a set of k? The answer here is also k. In each of the ways you are choosing, you are excluding one object (in other words, you are choosing one object not to choose- does this make sense?) So, this is the same as "k choose 1". It turns out that / k \ / k \ \ i / is equal to \ k-i / (can you see why?) Anyway, in this case, "k choose zero" and "k choose k" are k! k! __________ and ___________ , respectively. 0!(k-0)! k!(k-k)! This would pose a few problems if we defined 0!=0, since the denominator would be zero in that case. Also, we can figure out without calculations that the way to choose k objects out of a set of k is only one (we can pick them all in only one way). Hence, 0! has to be 1, so that the numerator and denominator of the above fractions will be equal. I think that 1!=1 follows directly the definition of factorial which is n!=n(n-1)(n-2)(n-3)...3*2*1 (I suppose I should have put this earlier in the message, but I assumed that you knew it.) Anyway, this is one reason we define 0!=1. I guess you could say that necessity dictates that we do so. I hope all this makes sense. If not, feel free to write back. Demetri- Dr, or something...
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