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Permutations, Combinations, Probability - Download Questions

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Math Expert
Joined: 02 Sep 2009
Posts: 39580
Re: P&C, Probability - Download [#permalink]

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25 Oct 2009, 22:12
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eresh wrote:
I have a question regarding the #37 question and the answer given.

37. 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

37. The best answer is A.
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=2520.

Now, it is not mentioned that a number cannot be repeated in for the 3rd digit. Say in the 2nd digit the number is 3 and so it is for the 3rd digit. The structure can be 233--, now for the 4th digit the option is 8 not 7.

Surely then the answer given is not right. But I don't know how to proceed further. Can someone please help? Or am I wrong about the porcess?

You are absolutely right. I'd say more there is no correct answer in choices. Surely the second and the third digits can be the same (in 3 cases out of 15) and in this case the number of possibilities for the 4th and 5th will be 8 and 7 and not 7 and 6.

So here is my solution:
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

Total=4*3*1*8*7+4*4*3*7*6=2688

P.S. eresh, can you please post other questions (you consider as worth of discussion) in PS or DS forums, as not many check this tree for particular problems, but rather for the tips and tutorials. You may even duplicate this question as there could be other opinions about it. Thanks.
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Joined: 05 Jul 2009
Posts: 181
Re: P&C, Probability - Download [#permalink]

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25 Oct 2009, 22:46
Bunuel, I believe your solution is absolutely correct.

So here is my solution:
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

Thinking of the approach, I understand that there are a total 15 combination of the 2nd and 3rd digit, of which in 3 combination the digits will be same. Therefore, there are 12 other possibilities where the digits are not same. The red part 4*3 also is equal to 12 but I could not figure out why it was 4*3 (My brains seems to stop working when I try to focus too much :D). Anyway, your answer seems correct and thank you for that.
Math Expert
Joined: 02 Sep 2009
Posts: 39580
Re: P&C, Probability - Download [#permalink]

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25 Oct 2009, 23:26
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eresh wrote:
Bunuel, I believe your solution is absolutely correct.

So here is my solution:
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

Thinking of the approach, I understand that there are a total 15 combination of the 2nd and 3rd digit, of which in 3 combination the digits will be same. Therefore, there are 12 other possibilities where the digits are not same. The red part 4*3 also is equal to 12 but I could not figure out why it was 4*3 (My brains seems to stop working when I try to focus too much :D). Anyway, your answer seems correct and thank you for that.

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.

Hope now it's clear.
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Posts: 181
Re: P&C, Probability - Download [#permalink]

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25 Oct 2009, 23:56
Bunuel wrote:

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.

Hope now it's clear.

Ohh, Ok I got it now
Manager
Joined: 22 Sep 2009
Posts: 216
Location: Tokyo, Japan
Re: P&C, Probability - Download [#permalink]

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14 Nov 2009, 20:37
This looks great. Going to study hard on these. Thank you very much
Manager
Joined: 24 Jul 2009
Posts: 73
Location: United States
GMAT 1: 590 Q48 V24
Re: P&C, Probability - Download [#permalink]

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16 Nov 2009, 01:59
Kudos to Amardeep Sharma.
Look at the number of downloads, 1576 & 1185, and yet very few kudos.
Intern
Joined: 03 Nov 2009
Posts: 5
Re: P&C, Probability - Download [#permalink]

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03 Dec 2009, 07:29
Kudos, i downloaded, and its great.

Also kudos to brunel, for coming up with that great explanation: brilliant.
Intern
Joined: 02 Jan 2010
Posts: 8
Re: P&C, Probability - Download [#permalink]

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13 Mar 2010, 00:57
Thanks... Thats very useful
Intern
Joined: 23 Feb 2010
Posts: 9
Re: P&C, Probability - Download [#permalink]

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16 Mar 2010, 09:29
Thank you!!! great resource
Intern
Joined: 17 Feb 2010
Posts: 14
Re: P&C, Probability - Download [#permalink]

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24 Mar 2010, 20:10
whts the difference between the 2 files?
Intern
Joined: 26 Mar 2010
Posts: 47
Re: P&C, Probability - Download [#permalink]

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02 Apr 2010, 08:28
Very useful- thanks for creating it!
Intern
Joined: 29 Mar 2010
Posts: 43
Re: P&C, Probability - Download [#permalink]

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29 Apr 2010, 10:17
Great !!! This Is exactly what I needed.

Posted from my mobile device
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Ros.
Nice Post + Some help + Lucid solution = Kudos

The greatest pleasure in life is doing what people say you cannot do | Great minds discuss ideas, average minds discuss events, small minds discuss people.
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Manager
Joined: 26 Feb 2010
Posts: 79
Location: Argentina
Re: P&C, Probability - Download [#permalink]

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06 May 2010, 06:49
thanks for sharing, I will take a look at them
Intern
Joined: 01 Apr 2010
Posts: 1
Schools: Carnegie Mellon, Berkeley, UCLA, Stanford, Princeton
Re: Permutations, Combinations, Probability - Download Questions [#permalink]

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07 May 2010, 14:27
Thanks for the quiz.
Manager
Joined: 16 Feb 2010
Posts: 225
Re: Permutations, Combinations, Probability - Download Questions [#permalink]

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08 May 2010, 14:09
thanks for the material !
Manager
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Re: Permutations, Combinations, Probability - Download Questions [#permalink]

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09 May 2010, 15:19
thanks for the material
Manager
Joined: 27 May 2008
Posts: 126
Re: Permutations, Combinations, Probability - Download Questions [#permalink]

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10 May 2010, 06:38
Thanks for the valuable post
Manager
Joined: 21 Jan 2010
Posts: 223
Re: Permutations, Combinations, Probability - Download Questions [#permalink]

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13 May 2010, 02:43
Thanks a lot. Very much appreciated.

What is the difference between the two files except formatting?
Intern
Joined: 29 Mar 2010
Posts: 43
Re: Permutations, Combinations, Probability - Download Questions [#permalink]

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13 May 2010, 09:38
Thanks I really need to work on probability this is of great help

Posted from my mobile device
_________________

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Ros.
Nice Post + Some help + Lucid solution = Kudos

The greatest pleasure in life is doing what people say you cannot do | Great minds discuss ideas, average minds discuss events, small minds discuss people.
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Intern
Joined: 05 Apr 2010
Posts: 4
Re: Permutations, Combinations, Probability - Download Questions [#permalink]

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15 May 2010, 22:33
thank you for sharing!!
Re: Permutations, Combinations, Probability - Download Questions   [#permalink] 15 May 2010, 22:33

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