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How many different numbers can we have by changing the position of the

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How many different numbers can we have by changing the position of the  [#permalink]

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New post 24 Feb 2015, 01:56
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C
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E

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Question Stats:

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How many different numbers can we have by changing the position of the digits of the number 718844?

a) 15
b) 90
c) 180
d) 360
e) 720
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Joined: 02 Sep 2009
Posts: 62542
Re: How many different numbers can we have by changing the position of the  [#permalink]

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New post 24 Feb 2015, 03:35
2
1
Awli wrote:
How many different numbers can we have by changing the position of the digits of the number 718844?

a) 15
b) 90
c) 180
d) 360
e) 720


THEORY

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:
\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\)

For example number of permutation of the letters of the word "gmatclub" is \(8!\) as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

According to the above, the number of arrangements of 6-digit number 718844, where 8's and 4's are repeated twice is 6!/(2!2!) = 180.

Answer: C.

Hope it's clear.
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Re: How many different numbers can we have by changing the position of the  [#permalink]

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New post 24 Feb 2015, 03:38
Bunuel wrote:
Awli wrote:
How many different numbers can we have by changing the position of the digits of the number 718844?

a) 15
b) 90
c) 180
d) 360
e) 720


THEORY

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:
\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\)

For example number of permutation of the letters of the word "gmatclub" is \(8!\) as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

According to the above, the number of arrangements of 6-digit number 718844, where 8's and 4's are repeated twice is 6!/(2!2!) = 180.

Answer: C.

Hope it's clear.


Very clear. Thank you very much Bunuel!
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Re: How many different numbers can we have by changing the position of the  [#permalink]

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New post 24 Feb 2015, 10:44
chetan2u wrote:
6!/2!2!=180
ans C



Thank you very much!
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Re: How many different numbers can we have by changing the position of the  [#permalink]

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New post 28 Feb 2015, 07:38
Awli wrote:
How many different numbers can we have by changing the position of the digits of the number 718844?

a) 15
b) 90
c) 180
d) 360
e) 720


Permutation in a group where there are repeated elements = n!/(p! * p2! *...*pn!); where p1, p2, .., pn are the number of times the element is repeated.
So, here required numbers = 6!/(2! * 2!) since we have 6 digits and 8 and 4 are each repeated twice.
= 180
Hence option C

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Re: How many different numbers can we have by changing the position of the  [#permalink]

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Re: How many different numbers can we have by changing the position of the   [#permalink] 02 Sep 2016, 23:37
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