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if the following terms apply

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if the following terms apply

by sanju09 » Tue May 24, 2011 1:45 am
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


[spoiler]Source: Eric's collection on BTG


OA A[/spoiler]
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by manpsingh87 » Tue May 24, 2011 3:39 am
sanju09 wrote: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


[spoiler]Source: Eric's collection on BTG


OA A[/spoiler]
-,-,-,-,-;
leftmost digit can be filled by any of the 2,4,6,8 in 4 ways;
second digit from the left can be filled by any of the 1,3,5,7,9 in 5 ways;
third digit from the left can be filled by any of the 3,5,7 in 3 ways;
fourth digit from the left can be filled by any of the 7 remaining numbers in 7 ways;
fifth digit from the left can be filled by any of the 6 remaining numbers in 6 ways;

hence required no. of ways= 4*5*3*7*6=2520 hence A
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by Frankenstein » Tue May 24, 2011 3:59 am
manpsingh87 wrote:
sanju09 wrote: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


[spoiler]Source: Eric's collection on BTG


OA A[/spoiler]

fourth digit from the left can be filled by any of the 7 remaining numbers in 7 ways;
fifth digit from the left can be filled by any of the 6 remaining numbers in 6 ways;

hence required no. of ways= 4*5*3*7*6=2520 hence A
Hi,
By saying this, you are effectively taking for granted that 2nd digit and 3rd digit are distinct. When, the 2nd and 3rd digit are equal, 4th digit can be chosen from 8 numbers and 5th digit from 7 numbers. For, the 3 equal cases(2nd digit and 3rd digit being equal), we need to add few cases.This can be done in 4.3.(8.7 - 7.6) = 168. This number added to 2520 gives 2688.
So, I think the OA is incorrect.

Cheers!
Last edited by Frankenstein on Tue May 24, 2011 4:35 am, edited 1 time in total.
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by Anurag@Gurome » Tue May 24, 2011 4:19 am
sanju09 wrote: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

[spoiler]Source: Eric's collection on BTG
OA A[/spoiler]
The leftmost digit is even implies it can be any digit from 2, 4, 6, 8 (that is 4 digits).
The second is odd implies it can any digit from 1, 3, 5, 7, 9 (that is 5 digits).
The third digit is a non even prime implies any digit from 3, 5, 7 (that is 3 digits).
The fourth digit can be any digit, which is not being used in the earlier 3 numbers (that is 7 digits, as 3 digits have been used before).
The fifth digit can be any digit, which is not being used in the earlier 4 numbers (that is 6 digits, as 4 digits have been used before).

Case 1: If the 2nd and 3rd digits are the same.

No. of 5 digit integers that can be formed = 4 * 3 * 1 * 8 * 7 = 672 (here 4th and 5th digits an be any digit from 8 and 7 digits respectively, as 2nd and 3rd digits are the same)

Case 2: If the 2nd and 3rd digits are not the same.

2nd digit can be any digit from 1 and 9
3rd digit can be any digit from 3, 5, and 7

No. of 5 digit integers that can be formed = 4 * 2 * 3 * 7 * 6 = 1,008

Case 3: If the 2nd and 3rd digits are from 3, 5, or 7.

2nd digit can be any digit from 3, 5, 7
3rd digit can be any digit from 3, 5, 7, but would not be the one that is already used as the 2nd digit

No. of 5 digit integers that can be formed = 4 * 3 * 2 * 7 * 6 = 1,008

Therefore, required no. of 5 digit numbers = 672 + 1,008 + 1,008 = 2,688, which is not there in the given answer choices.
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by djiddish98 » Tue May 24, 2011 4:28 am
This question seems to be worded poorly. Can the fourth and fifth digits be identical? When I read "two random digits not used before in the number" I assumed this was referencing just the first 3 digits, so they could be the same.
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by sanju09 » Wed May 25, 2011 1:27 am
djiddish98 wrote:This question seems to be worded poorly. Can the fourth and fifth digits be identical? When I read "two random digits not used before in the number" I assumed this was referencing just the first 3 digits, so they could be the same.
The Source Explanation is:


The best answer is [spoiler]A[/spoiler].

The first digit has 4 options (2,4,6,8 and not 0), the second has 5 options (1,3,5,7,9) the third has 3 options (3,5,7 and not 2), the fourth has 7 options (10-3 used before) and the fifth has 6 options (10-4 used before). The total is 4*5*3*7*6 = [spoiler]2520[/spoiler].
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by djiddish98 » Wed May 25, 2011 4:21 am
sanju09 wrote:
The first digit has 4 options (2,4,6,8 and not 0), the second has 5 options (1,3,5,7,9) the third has 3 options (3,5,7 and not 2), the fourth has 7 options (10-3 used before) and the fifth has 6 options (10-4 used before). The total is 4*5*3*7*6 = [spoiler]2520[/spoiler].
As Anurag pointed out, this explanation assumes a unique number in the second and third slot, although the question makes no statement that all numbers have to be unique. I assume the real GMAT problems won't be this ambiguous.
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by mmenifi » Wed May 25, 2011 5:52 am
Hi all,

I am just wondering why zero is not considered. Isn't zero an even number?
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by djiddish98 » Wed May 25, 2011 7:34 am
mmenifi wrote:Hi all,

I am just wondering why zero is not considered. Isn't zero an even number?
If a 5 digit number began with 0, it would technically be a 4 digit number, and out of the realm of possible numbers combinations that the question is asking for.

05,328 is really 5,328.
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