# How many odd numbers are there between 3000 and 7000

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

## Answers and Replies

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vsage
What do you mean no repeat? No repeated digits?

dextercioby
Homework Helper
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.

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xanthym
Omid said:
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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1008?

xanthym said:
{Valid Permutations Between 3000 & 7000} = 1008

~~
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

Lack of enough IQ :D

vsage said:
(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

(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.

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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

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

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.

The book is wrong then.
Thank you very much.

Healey01 said:
This I checked by brute force, not mathematics.
Sorry but, what do you mean "by brute force"?