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How many 4 digit numbers can be constructed from the numbers 1,2,3 that are odd and less than 2000.

My answer: 1(less than 2 so it's less than 2000) * 3 * 3 * 5 (odd numbers)

Why is this wrong ? : answer is 18 can someone explain why

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You mean $1\cdot 3\cdot 3\cdot 2$. Remember that the last digit must be $1$ or $3$. – André Nicolas Jun 27 '13 at 0:38
Oh right ! Sorry I missed that thanks – MethodManX Jun 27 '13 at 0:39
$33^2 = 1089$ fits your description: 4 digit number constructed from the numbers 1,2,3 that are odd and less than 2000. If you pose a sloppy question, you might get a sloppy answer. – wolfies Jun 27 '13 at 13:36

Hint: You're not allowed to use $5,7,9$.

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Your process seems to be right, but I think you picked the wrong values for your odd numbers.

A four digit number to answer your problem has the following form: $$\square\quad\square\quad\square\quad\square$$ Where each $\square$ could be a number in the set $\{1, 2, 3\}$.

However, we can further refine this: We need the number to be less than $2000$, so, as you correctly deduced, the first digit must be $1$: $$1\quad\square\quad\square\quad\square$$

We can further refine this by noting that the number must be odd. This means that the last digit must be $1$ or $3$. So, we can say that our number will look like: $$1\quad\square\quad\square\quad\{1, 3\}$$ The center two digits have no restriction, so they can be any of $\{1, 2, 3\}$. So our final number looks like: $$1\quad\{1, 2, 3\}\quad\{1, 2, 3\}\quad\{1, 3\}$$

To determine how many numbers fit this form, we multiply the count of possibilities for each digit. That is: $$\underbrace{1}_{\text{first digit}}\times\underbrace{3}_{\text{second digit}}\times\underbrace{3}_{\text{third digit}}\times\underbrace{2}_{\text{fourth digit}} = 18\;\text{possibilities}$$

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