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Intern  Joined: 02 Aug 2007
Posts: 1
How many 5 digit numbers can be created if the following  [#permalink]

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1
6 00:00

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(N/A)

Question Stats: 62% (02:54) correct 38% (01:09) wrong based on 292 sessions

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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

In my way:C(4,1)C(5,1)C(3,1)A(7,2)+C(4,1)C(5,1)C(3,1)A(8,2)

But there is no correct answer in my way...
Math Expert V
Joined: 02 Sep 2009
Posts: 65767
Re: How many 5 digit numbers can be created if the following  [#permalink]

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5
9
RACHITKH wrote:
I got 2688..

Following is my calculation:

First digit can be 2,4,6,8
2nd digit can be 1,3,5,7,9
3rd digit can be 3,5,7

Notice that 2nd and 3rd digit have same numbers thus there is a possibilty of repeat numbers. Only 1 and 9 are not repeated.
Thus 4*2*3*7*6 =1008 (This considers the case when 2nd digit is 1 or 9)
Now consider the second case where the second digit is 3, 5 or 7. In this scenario 2nd and 3rd digit can be same.
Thus we get 2 equations, one when the number repeats and the other when the number doesn't repeat
4*3*2*7*6 =1008 (2nd digit is 3,5,7 and 3rd digit is different than 2nd)
4*3*1*8*7 =672 (2nd digit is 3,5,7 and 3rd digit is same as 2nd)

2688 is the correct answer, but it's not in answer choices, so OA for this question is wrong.

This question was posted in another topic (permutations-combinations-probability-download-questions-57156-20.html), below is my solution from there:

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?

Note that the second and the third digits can be the same (in 3 cases out of 15).

1. second and third digits are the same:
4*3*1*8*7

2. second and third digits are not the same:
4*4*3*7*6
(Think about it this way: we don't want second and third digits to be the same, let's first choose the third digit (clearly it doesn't matter which one we choose first) how many possibilities are there? 3 (3 odd primes), than choose the second digit how many possibilities are there as one odd prime is already used? 5-1=4 --> 3*4=4*3.)

Total=4*3*1*8*7+4*4*3*7*6=2688.
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##### General Discussion
Intern  Joined: 01 Mar 2007
Posts: 2
Location: Seattle
Re: How many 5 digit numbers can be created if the following  [#permalink]

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I got 2688..

Following is my calculation:

First digit can be 2,4,6,8
2nd digit can be 1,3,5,7,9
3rd digit can be 3,5,7

Notice that 2nd and 3rd digit have same numbers thus there is a possibilty of repeat numbers. Only 1 and 9 are not repeated.
Thus 4*2*3*7*6 =1008 (This considers the case when 2nd digit is 1 or 9)
Now consider the second case where the second digit is 3, 5 or 7. In this scenario 2nd and 3rd digit can be same.
Thus we get 2 equations, one when the number repeats and the other when the number doesn't repeat
4*3*2*7*6 =1008 (2nd digit is 3,5,7 and 3rd digit is different than 2nd)
4*3*1*8*7 =672 (2nd digit is 3,5,7 and 3rd digit is same as 2nd)
Intern  Joined: 11 Sep 2013
Posts: 3
Location: India
Concentration: Marketing, Entrepreneurship
GMAT 1: 670 Q47 V35
GPA: 3.65
WE: Editorial and Writing (Journalism and Publishing)
Re: How many 5 digit numbers can be created if the following  [#permalink]

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no. of options when the 3rd & 2nd place are repeating - A - 4x3x3x8x7 - 2016
(as 3,5,7 are the only options for repeating numbers)
no. of options when the 3rd & 2nd place are NOT repeating - B - 4x2x3x7x6 - 1008
(1,9 for second place and 3,5,7 for thrid place)
Ans. A + B = 3024
Math Expert V
Joined: 02 Sep 2009
Posts: 65767
Re: How many 5 digit numbers can be created if the following  [#permalink]

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anchitg wrote:

no. of options when the 3rd & 2nd place are repeating - A - 4x3x3x8x7 - 2016
(as 3,5,7 are the only options for repeating numbers)
no. of options when the 3rd & 2nd place are NOT repeating - B - 4x2x3x7x6 - 1008
(1,9 for second place and 3,5,7 for thrid place)
Ans. A + B = 3024

It should be:
1. second and third digits are the same:
4*3*1*8*7

2. second and third digits are not the same:
4*4*3*7*6

Check here: how-many-5-digit-numbers-can-be-created-if-the-following-50488-20.html#p729286

Hope it helps.
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Non-Human User Joined: 09 Sep 2013
Posts: 15593
Re: How many 5 digit numbers can be created if the following  [#permalink]

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_________________ Re: How many 5 digit numbers can be created if the following   [#permalink] 05 Sep 2019, 06:47

# How many 5 digit numbers can be created if the following Question banks Downloads My Bookmarks Reviews Important topics   