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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]

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New post 13 Aug 2007, 05:10
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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...
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Re: How many 5 digit numbers can be created if the following  [#permalink]

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New post 25 May 2010, 03:07
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)
Adding 1008+1008+672= 2688
Hence the Answer


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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Re: How many 5 digit numbers can be created if the following  [#permalink]

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New post 24 May 2010, 21:44
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)
Adding 1008+1008+672= 2688
Hence the Answer
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Re: How many 5 digit numbers can be created if the following  [#permalink]

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New post 16 Sep 2013, 03:22
How about this as the answer -

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
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Re: How many 5 digit numbers can be created if the following  [#permalink]

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New post 16 Sep 2013, 03:40
anchitg wrote:
How about this as the answer -

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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Re: How many 5 digit numbers can be created if the following  [#permalink]

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New post 05 Sep 2019, 06:47
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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

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