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Intern  Joined: 08 Jan 2015
Posts: 25
How many different numbers can we have by changing the position of the  [#permalink]

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2 00:00

Difficulty:   15% (low)

Question Stats: 71% (00:54) correct 29% (01:15) wrong based on 100 sessions

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 62624
Re: How many different numbers can we have by changing the position of the  [#permalink]

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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.

Hope it's clear.
_________________
Intern  Joined: 08 Jan 2015
Posts: 25
Re: How many different numbers can we have by changing the position of the  [#permalink]

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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.

Hope it's clear.

Very clear. Thank you very much Bunuel!
Intern  Joined: 08 Jan 2015
Posts: 25
Re: How many different numbers can we have by changing the position of the  [#permalink]

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chetan2u wrote:
6!/2!2!=180
ans C

Thank you very much!
SVP  B
Joined: 06 Nov 2014
Posts: 1865
Re: How many different numbers can we have by changing the position of the  [#permalink]

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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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Posts: 14505
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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