Permutations

[Omid]

How many odd numbers are there between 3000 and 7000, NO REPEAT?
My answer is 1232 which is wrong according to the solutions manual.

[vsage]

What do you mean no repeat? No repeated digits?

[dextercioby]

Yes,otherwise the answer would have been very easy to give...


Daniel.

[vsage]

(might be good if he posted what he did so I can know what not to repeat)

Consider the case 3000-3999. Any number fitting the scheme above will be expressed as (3)(0, 1, 2, 4, 5, 6, 7, 8, 9)(0, 2, 4, 5, 6, 7, 8, 9)(7, 9) where I assumed 1 was the hundreds digit and 5 was the tens digit (so 1 was deleted from the selection of possible tens digits and 1 and 5 were deleted from the selection of unit and tens). Multiply the number above by 2 for the assumptions made about the units digit and by 1 for the assumptions made about the units digit to get the answer for that range 3000-3999. This should be the same answer for 5000-5999 (I get 288), and a similar method can be used to find the case for 4000-4999.

In case you're wondering about the pretty arbitrary way I wrote that, multiply the number of numbers inside each parenthesis to get the total number of permutations, and then multiply again by the number of choices I said I deleted for the units and tens digits.

I hope this made sense: I'm a little stumped how to explain it. Edit: I get a solution that's about 200 lower than what you listed.

[xanthym]

How many odd numbers are there between 3000 and 7000, NO REPEAT?
My answer is 1232 which is wrong according to the solutions manual.

For the moment, consider ALL 4 digit numbers, even those with 0 in the Thousand's position.
A number is odd if the One's digit is odd. Therefore, 5 possibilities exist for the One's digit {1, 3, 5, 7, 9}. The Ten's digit cannot repeat the One's, and thus 9 possibilities exist for it. The Hundred's digit cannot repeat either of the previous two, yielding 8 more choices. Finally, by the same logic, 7 choices remain for the Thousand's digit. Under these conditions, a total of (5x9x8x7) unique numbers would exist IF ALL 4 digit numbers were valid. However, we wish to consider only those with Thousand's digit {3, 4, 5, 6}. Thus, we include only (4/10) of the previous value, given by {5x9x8x7x4/10).

{Valid Permutations Between 3000 & 7000} = 1008


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[Omid]

{Valid Permutations Between 3000 & 7000} = 1008


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Your solution looks eager and completely new to me. But the answer in the book is this:
4 X 8 X 7 X 5 = 1120.
First let me clear it up:
How many odd numbers with distinc digits are there between 3000 and 7000?
Distinc means a number like 3011 is not valid because 1 is used two times.
Are you still sure of your answer?

Thanks

[Omid]

(might be good if he posted what he did so I can know what not to repeat)

Consider the case 3000-3999. Any number fitting the scheme above will be expressed as (3)(0, 1, 2, 4, 5, 6, 7, 8, 9)(0, 2, 4, 5, 6, 7, 8, 9)(7, 9) where I assumed 1 was the hundreds digit and 5 was the tens digit (so 1 was deleted from the selection of possible tens digits and 1 and 5 were deleted from the selection of unit and tens). Multiply the number above by 2 for the assumptions made about the units digit and by 1 for the assumptions made about the units digit to get the answer for that range 3000-3999. This should be the same answer for 5000-5999 (I get 288), and a similar method can be used to find the case for 4000-4999.

In case you're wondering about the pretty arbitrary way I wrote that, multiply the number of numbers inside each parenthesis to get the total number of permutations, and then multiply again by the number of choices I said I deleted for the units and tens digits.

I hope this made sense: I'm a little stumped how to explain it. Edit: I get a solution that's about 200 lower than what you listed.


Sorry but I don't get it :cry:

[Gamma]

(3 4 5 6) (0 1 2 3 4 5 6 7 8 9) (0 1 2 3 4 5 6 7 8 9) (1 3 5 7 9)

Within parenthesis are the individual digits of this 4 digit number.
For each digit in the first parenthesis, there are 8 ways of choosing the second digit (why? because you don't want to repeat the first or last digit)
Now, number of ways to choose the 3rd digit is 7 ( because you don't want to repeat first, last or the socond digit). Since there are 4 ways to choose the first digit and 4 ways to choose the last digit the permutation is
4 X 8 X 7 X 4.

Your book answer seems to have put 5 in the place of 4 in my equation. I don't know why they did it. or if I am missing some thing.

[Gamma]

I am finding a mistake in my solution. Mistake is in the number of ways to choose the first and last digits.

If the first digit is 3 last digit could be one of (1 5 7 9) = 4 possibilities
If the first digit is 4 last digit could be one of (1 3 5 7 9) = 5
If the first digit is 5 last digit could be one of (1 3 7 9) = 4
If the first digit is 6 last digit could be one of (1 3 5 7 9) = 5

Altogather we have 18 ways to choose the first and last digit.

So according to my understanding, answer is 18 X 8 X 7 = 1008

[Gamma]

I see that this is same answer as xanthym's solution.

[K.J.Healey]

1008 is the correct answer for all odd integers between 3000 and 7000 whose 1s,10s,100s,and 1000s digits are never equal. This I checked by brute force, not mathematics.

[Omid]

The book is wrong then.
Thank you very much.

[Omid]

This I checked by brute force, not mathematics.

Sorry but, what do you mean "by brute force"?