Permutations of 1234567890Date: 09/23/2001 at 06:09:19 From: Leeanna blyton Subject: Combinations I'm trying to find a pattern in combinations and how many combinations there are in 1234567890. I have found that 7 = 1 46 = 2 123 = 6 4567 = 24 I am now trying to find out how many combinations there are in 5-, 6-, 7-, 8-, 9-, and 10-digit numbers. Leeanna Date: 09/23/2001 at 07:19:34 From: Doctor Mitteldorf Subject: Re: Combinations Dear Leeanna, You're doing it just right. Doing examples with small numbers of digits will help you to see a pattern, and then you'll know how to calculate the combinations for any number of digits. (Actually, these are "permutations," not "combinations." For a review of Permutations and Combinations, see the Dr. Math FAQ: http://mathforum.org/dr.math/faq/faq.comb.perm.html ) The hard part about it is that the numbers get big so fast. You can list the permutations for 4 without too much trouble, but there are enough of them for 5 that you might have trouble listing them all without making mistakes. So the key is to come up with some kind of system so you're sure you have them all. Here's a suggestion for a system. Say you want to find the permutations of the digits 12345. Let's keep the 5 anchored at the end and list all the permutations that end in 5. Well, what you're left with is the digits 1234, and you can put them in any order. So you know how many there will be - you just did this problem, and found 24. Now let's count how many there are that end in 4. The other digits are 1235, and they can be in any order. But this is really the same as the problem you just did: there are 24 ways to order the numbers 1235, so there are 24 permutations of the digits 12345 that all end in 4. Now you're starting to see a pattern, I think. Follow the reasoning and you can count how many permutations there are going to be altogether without having to list them all. Once you've done this, you know the answer for 5 digits. Let's move on to 6. Can you apply the same kind of reasoning? How many permutations are there that end in 6? Well, you just calculated that by your astute and careful reasoning in the problem above. Then you list each of the other digits that the permutations could end in, and you have an idea how to get the total for 6 digits. Getting really abstract, you might formulate a rule: If I know how many permutations there are for N digits, then I can figure how many there are for N+1 digits. Describe the rule for how you do that. Make a chart, like the one you started, for the number of permutations of 1,2,3,4,5 6 and 7 digits. (Now I don't want you to peek. Finish the whole process that I've described above, and understand the patterns that you see. Only after you understand just what you've done and why it comes out the way it does, go to your dictionary and look up the word "factorial.") - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/