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# The 8 leaders of the G8 nations convene in Rome and stand in a row as

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The 8 leaders of the G8 nations convene in Rome and stand in a row as [#permalink]

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18 Aug 2015, 12:02
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61% (01:12) correct 39% (01:27) wrong based on 134 sessions

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The 8 leaders of the G8 nations convene in Rome and stand in a row as they get ready to have some pictures of them taken by the press. What is the probability that the picture that the New York Times' editors will randomly select for publishing the next day is one in which Berlusconi is not standing next to Obama? (assuming that there are pictures of all possible standing arrangements).

A. 1/8
B. 1/4
C. 1/2
D. 3/4
E. 7/8
[Reveal] Spoiler: OA

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The 8 leaders of the G8 nations convene in Rome and stand in a row as [#permalink]

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18 Aug 2015, 13:33
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reto wrote:
The 8 leaders of the G8 nations convene in Rome and stand in a row as they get ready to have some pictures of them taken by the press. What is the probability that the picture that the New York Times' editors will randomly select for publishing the next day is one in which Berlusconi is not standing next to Obama? (assuming that there are pictures of all possible standing arrangements).

A. 1/8
B. 1/4
C. 1/2
D. 3/4
E. 7/8

Glue those two together. We'd have 7 units: 1, 2, 3, 4, 5, 6, and {Berlusconi, Obama}. These 7 units can be arranged in 7! ways. Berlusconi and Obama within their unit can be arranged in 2! ways. Thus the total number of ways to arrange 8 leaders so that Berlusconi and Obama to stand together is 7!*2!.

8 people without any restriction can be arranged in 8! ways, thus in 8! - 7!*2! arrangements Berlusconi and Obama will not not be standing together.

$$P = \frac{8! - 7!*2!}{8!}=1 - \frac{7!*2!}{8!} = 1 - \frac{1}{4} = \frac{3}{4}$$.

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Re: The 8 leaders of the G8 nations convene in Rome and stand in a row as [#permalink]

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18 Aug 2015, 20:13
reto wrote:
The 8 leaders of the G8 nations convene in Rome and stand in a row as they get ready to have some pictures of them taken by the press. What is the probability that the picture that the New York Times' editors will randomly select for publishing the next day is one in which Berlusconi is not standing next to Obama? (assuming that there are pictures of all possible standing arrangements).

A. 1/8
B. 1/4
C. 1/2
D. 3/4
E. 7/8

Probability = $$\frac{Favorable cases}{Total no. of cases possible}$$

Total no. of cases = 8!

Let us consider the cases in which Obama and Berlusconi Stand next to each other :-

Consider Obama and Berlusconi as one single unit.
Thus in total there will be 7 members who are supposed to be arranged in 7 places = 7!
Obama and Berlusconi can interchange their places( i.e. OB or BO ) = 2 ways
Thus total no. of ways = 7! * 2

No. of cases in which O and B don't stand next to each other = Total no. of cases - No. of cases in which O and B stand next to each other
= 8! - (7! *2 )

Probability = $$\frac{8! - (7!*2)}{8!}$$
= 3/4
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Kudos [?]: 169 [0], given: 124

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Re: The 8 leaders of the G8 nations convene in Rome and stand in a row as [#permalink]

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02 Dec 2015, 13:07
total ways=8!

Consider 2 people standing together as 1 entity...so 7 people in 7! ways....n b/w these 2 people 2 ways..so 7!*2!
Thus
1−7!∗2!/8!=1−1/4=3/4

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Re: The 8 leaders of the G8 nations convene in Rome and stand in a row as [#permalink]

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22 Feb 2016, 20:28
oh man..tried to get it as fast as possible..and did not subtract from 1..picked 1/4...
although I did the right thing...
stick the two guys together -> arrange in 7!. these 2 can be arranged in 2!.
now this, divide by 8! of ways.
from total number of ways -> subtract the number of ways in which the guys sit together.
probability that they will sit together - 1/4
that they will not
1- 1/4 = 3/4

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Re: The 8 leaders of the G8 nations convene in Rome and stand in a row as [#permalink]

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28 Sep 2017, 09:29
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Re: The 8 leaders of the G8 nations convene in Rome and stand in a row as   [#permalink] 28 Sep 2017, 09:29
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