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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