combinatory : GMAT Problem Solving (PS)
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# combinatory

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Intern
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Joined: 02 May 2011
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20 Nov 2011, 10:56
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choose a five digit lock code in which the first e last digit must be ODD and aren’t allowed repetition:

the result is 8*7*6*5*4=6720 and until now is okay..

But is it possible to calculate this result in this way:?

1)calculate all the possible arrangements with 5 digits= 10!/5!
2) calculate all the numbers that have in the first and in the last place an EVEN number.

then subtract 1)-2) and you should have the result above..
why this reasoning is flawed?
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Joined: 23 Jun 2009
Posts: 361
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
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20 Nov 2011, 11:10
mushyyy wrote:
choose a five digit lock code in which the first e last digit must be ODD and aren’t allowed repetition:

the result is 8*7*6*5*4=6720 and until now is okay..

But is it possible to calculate this result in this way:?

1)calculate all the possible arrangements with 5 digits= 10!/5!
2) calculate all the numbers that have in the first and in the last place an EVEN number.

then subtract 1)-2) and you should have the result above..
why this reasoning is flawed?
Because it gives permission to repetition of numbers.
Intern
Status: student
Joined: 02 May 2011
Posts: 24
Location: Italy
Followers: 0

Kudos [?]: 48 [0], given: 7

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20 Nov 2011, 11:31
maliyeci wrote:
mushyyy wrote:
choose a five digit lock code in which the first e last digit must be ODD and aren’t allowed repetition:

the result is 8*7*6*5*4=6720 and until now is okay..

But is it possible to calculate this result in this way:?

1)calculate all the possible arrangements with 5 digits= 10!/5!
2) calculate all the numbers that have in the first and in the last place an EVEN number.

then subtract 1)-2) and you should have the result above..
why this reasoning is flawed?
Because it gives permission to repetition of numbers.

sorry but where?
10!/5! is like ABCDEXXXXX
where A,B,C,D,E are 5 different numbers choosen in a gropu of ten..
Senior Manager
Joined: 23 Jun 2009
Posts: 361
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
Followers: 5

Kudos [?]: 130 [0], given: 80

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20 Nov 2011, 11:44
mushyyy wrote:
maliyeci wrote:
mushyyy wrote:
choose a five digit lock code in which the first e last digit must be ODD and aren’t allowed repetition:

the result is 8*7*6*5*4=6720 and until now is okay..

But is it possible to calculate this result in this way:?

1)calculate all the possible arrangements with 5 digits= 10!/5!
2) calculate all the numbers that have in the first and in the last place an EVEN number.

then subtract 1)-2) and you should have the result above..
why this reasoning is flawed?
Because it gives permission to repetition of numbers.

sorry but where?
10!/5! is like ABCDEXXXXX
where A,B,C,D,E are 5 different numbers choosen in a gropu of ten..

But the formula you are giving is not the formula of combination it is the formula of permutation.
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20 Nov 2011, 11:54
Quote:
But the formula you are giving is not the formula of combination it is the formula of permutation.

sorry but I can't understand where my reasoning is wrong:

I'm studying on the mgmat book and they say "when you choose x person in a group of y, then the formula is Y!/X!"(the order is important):

123XXXX= 7!/4!= 210 ways

if I want to choose 5 numbers in a group of ten, in which the order is important, is wrong 10!/5!?

12345XXXXX

if I'm wrong, could you write the right formula if I want to calculate the result in "my " way?
Senior Manager
Joined: 23 Jun 2009
Posts: 361
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
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Kudos [?]: 130 [0], given: 80

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20 Nov 2011, 12:12
mushyyy wrote:
Quote:
But the formula you are giving is not the formula of combination it is the formula of permutation.

sorry but I can't understand where my reasoning is wrong:

I'm studying on the mgmat book and they say "when you choose x person in a group of y, then the formula is Y!/X!"(the order is important):

123XXXX= 7!/4!= 210 ways

if I want to choose 5 numbers in a group of ten, in which the order is important, is wrong 10!/5!?

12345XXXXX

if I'm wrong, could you write the right formula if I want to calculate the result in "my " way?

You are selecting the group w/o order. i.e. 75631 is different from 76513. But the number groups are same.
Re: combinatory   [#permalink] 20 Nov 2011, 12:12
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