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How many three digit numbers can be formed from the digits $1,2,3,4,5$ and $6$, if each digit can only be used once? How many of these are odd numbers? How many are greater than $330$?

I've determined that the total number of combinations formed by $3$ digits is $$6\times 5\times 4$$

Now, for determining how many of these combinations are odd, each of the combinations must end with $1,3$ or $5$. The solution shows that the correct answer is $$3\times(5\times4)$$

I'm curious whether this is correct because there are $6$ choices for the first number, not $5$, so wouldn't the answer be $6\times5\times3$?

I also thought to myself, what if the first number chosen was a $1$ and the second number chosen was a $3$, then for the last digit there is only $1$ option $(5)$ since $1$ and $3$ have already been used, so there is no way my solution could be correct.

So maybe the answer is $3 \times 2 \times 3$ since there are: $3$ options for the first number $(2,4,6)$, $2$ options for the second number $(4,6)$ and $3$ options for the last number $(1,3,5)$.

OR

I think this is the logic that whoever wrote the solution was using... If you make sure to always leave one of the odd numbers available: There are $5$ choices for the first digit (all but $1$), $4$ choices for the second digit ... and I don't know how to proceed from here.

First choice possibilities (Assuming we leave the $1$ available) : $2,3,4,5,6$ - $5$ Options
Second choice possibilities (Still not touching the $1$): $5 - 1$ Options
Third choice possibilities : $1$ option (the $1$)

Then repeat this for $1, 3$ and $5$. So $$3 \times (5\times4)$$

Can someone set me straight on this? Thanks for any help ahead of time.

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To see that the number of odd numbers is equal to $3\cdot(5\cdot4)$ instead of $6\cdot5\cdot3$ think as follows.

  1. Fix an odd number in the last position, f.e. digit $3$ (so start from the last digit and not from the first).
  2. Now you have $5$ remaining digits to place on the first position and after that $4$ remaining digits to place on the second position. This gives you $5\cdot4$ possible numbers that end with $3$.
  3. Repeat for digit $1$ and digit $5$, so that you have in total $$5\cdot4+5\cdot4+5\cdot4=3\cdot (5\cdot4)$$ possible odd numbers.
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  • $\begingroup$ Wow, thanks. This really cleared it up for me! $\endgroup$ – Riptyde4 Dec 14 '14 at 18:29
  • $\begingroup$ Would you say a generic rule for approaching a problem like this is to begin from the position upon which there are limitations placed? $\endgroup$ – Riptyde4 Dec 14 '14 at 18:31
  • $\begingroup$ @Riptyde4. Yes, nice remark. This is exactly the rule to use, plus to separate the process in steps and use the multiplication principle. As you said, the first step should be the one with the most limitations. $\endgroup$ – Jimmy R. Dec 14 '14 at 18:35
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A number is odd if it ends in 1, 3 or 5, so we have to pick that first so for a 3 digit number we have

[][][3]

where the number inside the box is how many choices for a digit at that position we have. Then, since we have picked one for the last box, there are 5 numbers left

[5][][3]

and since we now have 4 numbers left

[5][4][3]

To get our answer multiple out the numbers in the boxes

5*4*3 or 3*(5*4) = 60

Another way of thinking about this is that since half of the numbers we can pick from are odd and half are even, that combinations produce an odd number is half off the total combinations (6*5*4)/2 = 3*5*4

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