Hi!
I would appreciate your answer to the following question:
In permutations, I don't understand why the answers always start calculating from the first letter\digit and why the answer is not the same if we start calculating the possibilities from a different letter\digit?
For example, when calculating the number of three digit numbers in which the digits are different, the right answer is:
9 (every digit besides zero) * 9 (every digit besides the first) * 8 (every digit besides the first two) = 648
However, if I want to start calculating from the last digit, the answer is:
10 (every digit) * 9 (every digit besides the first) * 7 (every digit besides zero and the first two) = 630.
Why isn't the answer identical?
Thanks!
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- LiranE752
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- danr969
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Re: order in permutations
the second way you computed the number of possibilities doesn't preclude having a zero in the most significant digit.
- danr969
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Re: order in permutations
more precisely, to count starting from the right, you can split the cases as to whether zero appears in the two right digits or not.
First, assume that zero never appears at all. Then, there are 9 * 8 *7 possibilities. Next, if zero appears in the ones digit, there are 9 possibilities for the tens digit and 8 possibilities for the hundreds digit. 72 more possibilities. Finally, zero could be in the middle digit.
This leaves another 72 possibilities.
9*8*7 + 72 + 72 again gives 648, same as counting directly from the left. The constraint seems easier to account for when starting from the left. Counting from the right alters the possibilities depending on whether zero has been used already.
First, assume that zero never appears at all. Then, there are 9 * 8 *7 possibilities. Next, if zero appears in the ones digit, there are 9 possibilities for the tens digit and 8 possibilities for the hundreds digit. 72 more possibilities. Finally, zero could be in the middle digit.
This leaves another 72 possibilities.
9*8*7 + 72 + 72 again gives 648, same as counting directly from the left. The constraint seems easier to account for when starting from the left. Counting from the right alters the possibilities depending on whether zero has been used already.
- RonPurewal
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Re: order in permutations
Very nice summary above.
In terms of the products you're already using, the above gives 9*8*7 + 1*9*8 + 9*1*8.
The order here is (possibilities for ones digit) * (possibilities for tens digit) * (possibilities for hundreds digit).
In terms of the products you're already using, the above gives 9*8*7 + 1*9*8 + 9*1*8.
The order here is (possibilities for ones digit) * (possibilities for tens digit) * (possibilities for hundreds digit).
- LiranE752
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Re: order in permutations
Yeah. Got it.
Thanks a lot!
Thanks a lot!
- danr969
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Re: order in permutations
you could also find the answer by first ignoring the constraint.
This gives 10 * 9 * 8 = 720 possibilities, some of which start with zero. Then, subtract the 72 possibilities that start with zero.
This again leaves 648 possibilities.
This gives 10 * 9 * 8 = 720 possibilities, some of which start with zero. Then, subtract the 72 possibilities that start with zero.
This again leaves 648 possibilities.
- RonPurewal
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Re: order in permutations
danr969 Wrote:you could also find the answer by first ignoring the constraint.
This gives 10 * 9 * 8 = 720 possibilities, some of which start with zero. Then, subtract the 72 possibilities that start with zero.
This again leaves 648 possibilities.
That works, too.
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