Pascal's Triangle and the ice cream conesDate: Fri, 9 Dec 1994 11:18:27 -0800 (PST) From: Moore Classroom Account Subject: math Dear Dr. Math, We have a new problem for you from Mrs. Moore's class. Me (Gavin) and Andrew, we'll explain it to you. There are 31 flavors of ice cream at the local ice cream store. How many triple scoop cones can you make? Remember that two chocolate on the bottom and one vanilla on the top is the same as two chocolate on the top and one vanilla on the bottom. Now (I Gavin) we'll explain what I found. #flavors #scoops 1 1 2 4 3 10 4 20 5 35 What I found was really interesting. I found that 3 times 3 equals 9 plus one would be 10. That 1 is the first number in the scoops column. 4 times 4 is 16, plus that 4 equals twenty. And that twenty is the second number in the scoops columm. This works the whole way but Mrs. Moore said that we cannot write that as a function, so can you help us? The pattern was also found on Pascal's triangle. Date: Fri, 9 Dec 1994 15:57:14 -0500 (EST) From: Dr. Ken Subject: Re: math Hello there! This is a very nice problem. We came up with two different ways to get the answer, both of which are the same. So that's a relief. Solution 1: You noticed that the pattern shows up on Pascal's Triangle. So we can use that information to see exactly what the function should be. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 Notice that the nth term in your sequence appears in the (n+2)nd row, and in the (n-1)st place in that row. Note that the top row in the triangle is called the zeroth row, and the leftmost entry is called the zeroth place in the row. So since the entries in Pascal's triangle are given by "P choose K", the nth term in your sequence is "(n+2) choose (n-1)." The way we find its value is by the relation P! "P choose K" = --------- K!*(P-K)! So we have (n+2)! (n+2)(n+1)(n) --------------------- = --------------- , which is the first formula. (n-1)!*((n+2)-(n-1))! 6 Solution 2: Let's think about the number of ways you can choose the flavors, and divide it into cases. Then we'll add to get the total number of ways. Case 1: all three scoops are the same It looks like we'll be able to have as many different configurations as there are different flavors, so this term is just n. Case 2: all three scoops are different The number of ways to choose three different flavors from a set of n flavors is "n choose three", i.e. n!/(3!*(n-3)!) = (n)(n-1)(n-2)/6. Case 3: two scoops the same, one scoop different There will be n ways to choose the flavor that has two scoops in our cone, and then (n-1) ways to choose the other flavor (we can't choose the one we've already used for the other two scoops!). So this is (n)(n-1). These are all the possibilities that we can get. So we add them up, and find that the total number of ways to choose these scoops is n + n(n-1) + n(n-1)(n-2)/6. You can verify algebraically that this is the same as in the first way. I hope this helps you see what's going on! Write back if you have more questions. -Ken "Dr." Math & Ethan "Pascal's" Triangle |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/