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Kisses

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Kisses

by Uri » Fri Apr 17, 2009 1:35 am
Six three-female-representative delegations attend a conference. The representatives give a double kiss on a check when they are introduced to one another. How many kisses on a check are possible if each delegate gives a double kiss on a check only once with every other attendant except with those of her delegation?

(A) 766
(B) 270
(C) 180
(D) 144
(E) 72

I have come across this question in an online community. Till now I don't know the official answer. Can you please show the correct procedure?
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by m&m » Fri Apr 17, 2009 4:57 am
so we have 6 teams of 3

a1, a2, a3
b1, b2, b3
...
f1, f2, f3

Team a
a1 cannot kiss a2, or a3 but can kiss everyone else, so 6*3 - 3 = 18-3=15
same with a2 and a3

so team a can kiss 15 ppl each so 15+15+15

Team b
cannot kiss other members of team b, and has already kissed team a so each member can only kiss 12 other people so 12+12+12

notice the pattern, each member of the team kisses the same number of ppl but with each passing team there are 3 less people....

15*3
12*3
9*3
6*3
3*3

once the last 2 teams have kissed each other the last team cannot kiss each other so we add up the above = 45*3 = 135

Don't forget that there are 2 kisses per cheek so 135*2 = 270 so B

hope that helps, it looks long above but it was short on paper
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by shibal » Fri Apr 17, 2009 2:41 pm
i tought it this way:

18 girls in total
any girl will kiss 15 different girls (since she won't kiss herself and the 2 girls from her team).
there are 18 girls in total so this will happen to all of them
18*15=270...
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by m&m » Fri Apr 17, 2009 3:25 pm
Shibal, though your soln gives you the right answer I don't think the method is right. First off you don't account for the number of kisses, had they kissed each other 3 times then your answer would be off though your method is the same. Also you don't account for 2 people kissing (however many kisses) more than once.

Though your method is quick and easy I don't think it would apply if the question was changed. Think you got lucky with the numbers.
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by Uri » Sat Apr 18, 2009 11:58 pm
Thanks to both of you for the solution and the discussion.

Finally I have got the OA. It's 270
The OE is given below.
All= 2* C(18,2)
Prohibited= 2* {6* C(3,2)}
Thus, permitted no of kisses= 2* C(18,2) - 2* {6* C(3,2)}= 270

But the OE seems a bit confusing to me. Should not we multiply "All" and "Prohibited" by 2 once again, as per the technique? C(18,2) represents the number of possible combinations; this is very easy to understand. And it is multiplies by 2 to account for the double kisses. But each member in a combination or group gives one double kiss. So, in that case there are 4 kisses in a group. This part in the OE is confusing me. Can you please identify where I am making mistake?
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