Numbers with the Digit 3Date: 10/27/1999 at 21:58:36 From: Namita Subject: Permutations and combinations The question that I'm having trouble with is this: In how many numbers between 1000 and 9999 does the digit 3 occur? I know that each number must be at most four digits, and that 3 can occur in all four places (units, tens, hundreds, and thousands), but there is no possible way to count them all. I can't figure out what formula could help me to solve this - am I looking for combinations or permutations? I'll appreciate any help you can give. Thank you, Namita Date: 10/28/1999 at 03:49:16 From: Doctor Mitteldorf Subject: Re: Permutations and combinations Dear Namita, There are no rules or formulas for problems like this. It's good to get used to the idea that in statistics problems and combinations problems you always want to go back to first principles and try to get a picture inside your head of what it is you're counting. One way to do this is just to write a computer program to do the counting for you. It's an easy program to write if you know Pascal or Basic or C, and will take only a few minutes. It's a legitimate way to check your other method. We'll have to invent the other method. Start with all the 3xxx numbers. There are just 1000 of them. Then, how many numbers have 3 in the second digit? 100 each for 13xx, 23xx, 33xx, 43xx, 53xx, 63xx, 73xx, 83xx and 93xx. But we've already counted the 33xx's when we counted the 3xxx's. So we have 800 new ones. Can you continue with this kind of reasoning? It does get a bit more complicated when you have to think about what's excluded (because you've already counted it) in the steps that follow. The final answer I get is 3168. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
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