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How many 5-digit positive integers exist the sum of whose

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How many 5-digit positive integers exist the sum of whose [#permalink] New post 05 Aug 2006, 05:19
How many 5-digit positive integers exist the sum of whose digits are odd?
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Re: Comb problem [#permalink] New post 05 Aug 2006, 11:23
apollo168 wrote:
How many 5-digit positive integers exist the sum of whose digits are odd?


Possible combinations
1.ooooo
2.eeooo
3.eeeeo

1. 5^5
2. [5!/(2!*3!)]*5^5 - 5^4 (-5^4 to remove numbers start with 0)
3. [5!/4!]*5^5 - 5^4 (-5^4 to remove numbers start with 0)

Hence, 5^5 + 49*5^4 + 24*5^4
= 5^4(5+49+24) = 78*5^4 = 48750.
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 [#permalink] New post 05 Aug 2006, 17:33
Care to explain more?
I used a simpler method to get a different answer.
There are 3 out of possible 6 combinations of odd and even digits, that result in an odd sum.

There are 99999-10000 = 89999 5 digit numbers totally.
So , the number of odd sum 5 digit nos are 89999/2 or 45000
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 [#permalink] New post 05 Aug 2006, 21:36
ArvGMAT wrote:
Care to explain more?
I used a simpler method to get a different answer.
There are 3 out of possible 6 combinations of odd and even digits, that result in an odd sum.

There are 99999-10000 = 89999 5 digit numbers totally.
So , the number of odd sum 5 digit nos are 89999/2 or 45000

Sorry for the late respond. I had to go out and grab somethin to eat.

Good point.. you are absolutely right! Thanks!

I made silly mistake
Possible combinations
1.ooooo
2.eeooo
3.eeeeo

1. 5^5
2. [5!/(2!*3!)]*5^5 - [4]*5^4 ( to remove numbers start with 0)
3. [5!/4!]*5^5 - [4]*5^4 ( to remove numbers start with 0)

Hence, 5^5 + (50-4)*5^4 + (25-4)*5^4
= 5^4(5+46+21) = 78*5^4 = 45000

As you said.. there are equal numbers of those integers..
Easiest it can be = (99999-10000 + 1)/2 = 45000
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 [#permalink] New post 05 Aug 2006, 22:07
Total 5 digit numbers = 99999-9999 = 90000

Half of these will have odd sum of digits and half will have even sum of digits.
Answer = 90000/2 = 45000.
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 [#permalink] New post 06 Aug 2006, 07:25
freethinking, could you exlain how you did the following:

2. [5!/(2!*3!)]*5^5 - [4]*5^4 ( to remove numbers start with 0)
3. [5!/4!]*5^5 - [4]*5^4 ( to remove numbers start with 0)


number of ways

1)ODD1, ODD, ODD, ODD, ODD

2)ODD1, ODD, ODD, EVEN, EVEN

3)ODD1, EVEN, EVEN, EVEN, EVEN


1) can be accomplished in 5^5 ways

n1 = (5)^5

2) ODD1 can be chosen in 5 ways. Then we have to choose 2 ODD numbers and two EVEN numbers. They all can be choosen in 5 ways each and they can be placed in 4! ways.

n2 = (4!)*(5)^5 = 51000.

I know i commmitted an error here. I haven't subtracted the repititive combinations here. Could somebody help?

3) same thing can be done with this case.

somebody, please explain.

3) OD
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  [#permalink] 06 Aug 2006, 07:25
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How many 5-digit positive integers exist the sum of whose

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