Hello everyone, I have a little problem in permutation. Here's the problem: From the sets of numbers: 2,3,4,5,6,7,9, we want to create a number with a criteria:

The number is consist at least three digits

The number is less than 400

The number must be even

All the digits are different, i.e you can create 234 but cannot create 244

By using the methods of counting, I got 30 as an answer. But, is it true? And, if I want to use permutation formula to solve this problem, how can I apply it to get the same answer with the methods of counting? Thanks

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TopNewestThe answer is actually 25, I believe. Clearly, the hundreds place must contain a \(2\) or a \(3\). If it contains a \(2\), we can put either a \(4\) or a \(6\) in the units place and then any of the five remaining numbers in the tens place. This gives \(10\) possibilities. If the hundreds place contains a \(3\), then we can fill in the units place with a \(2\), a \(4\), or a \(6\) and the tens place with any of the five remaining numbers. This gives \(15\) possibilities. \(10 + 15 = \fbox{25}\). I think you counted the numbers which begin and end in \(2\), which are not allowed but would give an answer of \(30\) if they were included.

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Yeah I think your's is right.

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Considering the conditions, you may form one-digit numbers, two-digit number, or three-digit numbers. 1) For one-digit numbers, you may choose from 2, 4, or 6. Which means there are 3 possible numbers. 2) For two-digit numbers, you may only choose from either 2,4, or 6 to be your units digit. Which means you have 3 choices. For the tens digit, since a digit has been chosen as your ones digit, there will only be 6 choices left. Which means there will be 6

3 or 18 two digit numbers. 3) For the three-digit numbers. You only have two numbers as your choice for the hundreds digit since the numbers have to be less than 400. These are 2 or 3. However, choosing either will affect the number of possible even digits for the ones. So, it will be broken down into two cases: first, a number that starts with 2 and second, a number that starts with 3. For the first case since the first digit is 2, there will be two choices for the ones digit (4 or 6). Therefore, there will only be five choices for the tens. Hence, there will be 152 or 10 numbers. For the second case since the first digit is 3, there will be three choices for the ones digit (2, 4, or 6), and since two digits have been chosen for the hundreds (3) and ones (2,4, or 6), there will only be five choices for the tens. Therefor there will be 15*3 or 15 numbers. 4) All in all there will be 3+18+10+15 or 46 numbers. If you list down the numbers they are: 2,4, and 6 for a single-digit number. 24, 26, 32, 34, 36, 42, 46, 52, 54, 56, 62, 64, 72, 74, 76, 92, 94, and 96 for the two-digit numbers. 234, 236, 246, 254, 256, 264, 274, 276, 294, and 296 that starts with 2. 324, 326, 342, 346, 352, 354, 356, 362, 364, 372, 374, 376, 392, 394, and 396 that starts with three.Log in to reply

Sorry, some typo errors. 63 is supposed to be 6

3, 152 is supposed to be 152, and 153 should be 153. Thanks!Log in to reply

Ooops! Again! 63 is 6 times 3, 152 is 1 times 5 times 2. and 15*3 is 1 times 5 times 3. There is a problem with the symbol for asterisk. I don't know why. Thanks again.

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Comment deleted Jul 18, 2013

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In this case both hundredth place and ones place can have 2 which is not allowed as Sotiri said so, I think 25 is the answer.

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