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How many 5 digit numbers can be created if the following

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How many 5 digit numbers can be created if the following [#permalink] New post 16 Sep 2005, 16:44
How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?

a) 2520
b) 3150
c) 3360
d) 6000
e) 7500
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 [#permalink] New post 16 Sep 2005, 17:03
i get B

the leftmost number is even. so we have 5 choices (they didn't say anything about not using 0)
the second number is odd, again we have 5 choices
third is a non even prime number, ie 3 choices (3,5,7)
fourth is digit that is not used before so total of 7 choices
and fifth, not used before so 6 chices
therefore, 5*5*3*7*6=3150
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 [#permalink] New post 16 Sep 2005, 17:59
I get A) 2520

Method same as need700+ but I think zero should not be considered as the left most. So left most has 4 choices.
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 [#permalink] New post 16 Sep 2005, 23:08
Zero should not be considered as a starting digit in cases like this.

Only allow starting zeros where the question says or the context clearly requires (for instance in safe combinations).
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 [#permalink] New post 19 Sep 2005, 03:17
The OA is A...richardj is correct. You should not include zero as an option for the first digit.

I fell victim to this very mistake. It is frustrating because the problem asks for all even digits for the first digit, so I included zero. I got a problem wrong a week ago for not including zero when it asked for all even digits. What gives man!
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 [#permalink] New post 19 Sep 2005, 04:06
I agree A is the best of the possible answers here, but I do not think it is correct. Please tell me if I am wrong:

If we proceed with 4*5*3*7*6 we are not considering the case when the second and third digit are the same.

My solution would be:

(4*5*3-4*3)*7*6 + 4*3*8*7 = 2688

-4*3 because out of the 4*5*3 possibilities, 4*3 of them have the second and third digit equal.
4*3*8*7 because when the second and third digit are equal you have 8 different numbers that have not been used before.
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 [#permalink] New post 19 Sep 2005, 10:51
jdtomatito wrote:
I agree A is the best of the possible answers here, but I do not think it is correct. Please tell me if I am wrong:

If we proceed with 4*5*3*7*6 we are not considering the case when the second and third digit are the same.

My solution would be:

(4*5*3-4*3)*7*6 + 4*3*8*7 = 2688

-4*3 because out of the 4*5*3 possibilities, 4*3 of them have the second and third digit equal.
4*3*8*7 because when the second and third digit are equal you have 8 different numbers that have not been used before.


Why do we need to remove numbers where 2nd 3rd digits are same? The question didnt ask it..did it?

2nd digit can be one of 5 (1, 3, 5, 7, 9)
3rd didit can be one among (3, 5, 7)

there can be 5 * 3 = 15 arrangements between them. and all of these are DIFFERENT. So we need to include ALL of them.
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 [#permalink] New post 19 Sep 2005, 11:53
duttsit wrote:
jdtomatito wrote:
I agree A is the best of the possible answers here, but I do not think it is correct. Please tell me if I am wrong:

If we proceed with 4*5*3*7*6 we are not considering the case when the second and third digit are the same.

My solution would be:

(4*5*3-4*3)*7*6 + 4*3*8*7 = 2688

-4*3 because out of the 4*5*3 possibilities, 4*3 of them have the second and third digit equal.
4*3*8*7 because when the second and third digit are equal you have 8 different numbers that have not been used before.



Why do we need to remove numbers where 2nd 3rd digits are same? The question didnt ask it..did it?

2nd digit can be one of 5 (1, 3, 5, 7, 9)
3rd didit can be one among (3, 5, 7)

there can be 5 * 3 = 15 arrangements between them. and all of these are DIFFERENT. So we need to include ALL of them.


Duttsit,
If second and third digits are the same then, only two numbers are used up till we reach fourth digit, so there is more possibility for 4 and 5 th digits that is why it has to be considered seperately.

I agree with jdotmotio...

60 possibilty for first three digit combo, out of that 12 will have second and third digit repeated so

12*8*7 + 48*7*6 = 2688 should be the answer to tis question or the question should say "no digit can be repeated"

-Ranga
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 [#permalink] New post 19 Sep 2005, 12:24
ranga41 wrote:

Duttsit,
If second and third digits are the same then, only two numbers are used up till we reach fourth digit, so there is more possibility for 4 and 5 th digits that is why it has to be considered seperately.

I agree with jdotmotio...

60 possibilty for first three digit combo, out of that 12 will have second and third digit repeated so

12*8*7 + 48*7*6 = 2688 should be the answer to tis question or the question should say "no digit can be repeated"

-Ranga


Got your drift. Thanks Ranga.
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 [#permalink] New post 19 Sep 2005, 12:34
Ten Thousands Digit Possibilities: (2,4,6,8) = 4
Thousands Digit Possibilities: (1,3,5,7,9) = 5
Hundreds Digit Possibilities: (3,5,7) = 3
Tens Digit Possibilities: 10-3 = 7
Ones Digit Possibilities: 10-4 = 6

4*5*3*7*6 = 2520

Answer is A?
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 [#permalink] New post 19 Sep 2005, 12:43
Yeah, I see the problem.

If digits 2 and 3 are the same, there is an additional possibility for each of digits 4 and 5. So the cases would need to be considered separately.
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 [#permalink] New post 21 Sep 2005, 02:52
need700+ wrote:
i get B

the leftmost number is even. so we have 5 choices (they didn't say anything about not using 0)
the second number is odd, again we have 5 choices
third is a non even prime number, ie 3 choices (3,5,7)
fourth is digit that is not used before so total of 7 choices
and fifth, not used before so 6 chices
therefore, 5*5*3*7*6=3150


I really need someone to help me on thins:
This is how I calculated :
- 0 cannt be there in left most place. The no will become 4 digit in that case.
- Case 1:
Second and Third digit have diff number:
4C1*5C1*2C1 (7C1*6C1)


- case 2:
Both are same digits :
4C1*5C1*1 (8C1*7C1)
This is becoz only 2 digits have been used already.


And finally sum these 2.

40*42+20*56
=40(42+26)=2640
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Re: How many 5 digit numbers can be created if the following [#permalink] New post 20 Sep 2012, 00:40
The first place may be filled by 2,4,6,8 = 4
Second by 1,3,5,7,9 = 5
Third by 3,5,7 = 3
So, the total no. of combns for the three places are 4*5*3 = 60.
Given there are no restrictions on balance two places, they may be filled by 10 digits each.
Hence the maximum no. of combinations turns out to be 60*10*10 = 6000. However, there are restrictions, hence options d & e are ruled out.
Also, the answer has to be a multiple of 60, it rules out the option b as well.
Again, provided all digits of the number are different, fourth place can be filled by 7 and the fifth by 6 digits. So, in this case the number of numbers will be 60*7*6 = 60*42 = 2520.
However, the restriction on digits not being repeated is only on fourth and fifth place, the number of numbers created will be more than this.
So, by this methodology, I arrive at the answer of 3360 numbers.
Tell me if I am wrong.
Re: How many 5 digit numbers can be created if the following   [#permalink] 20 Sep 2012, 00:40
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