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# Delegations

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VP
Joined: 06 Feb 2007
Posts: 1023
Followers: 20

Kudos [?]: 131 [0], given: 0

Delegations [#permalink]  23 Mar 2007, 11:44
Please provide your explanations. Does anybody know a book that explains combinatorics/probability concepts well?

Six three-representative delegations attend an international conference. The representatives shake hands when they are introduced to one another. How many handshakes are possible if each delegate shakes hands only once with every other attendant except with those of his/her delegation?

A) 36
B) 72
C) 90
D) 135
E) 388
Manager
Joined: 25 May 2006
Posts: 227
Followers: 1

Kudos [?]: 19 [0], given: 0

D) 135

Lets say we got:
AAA BBB CCC DDD EEE FFF

1) For A's: 3*15=45 hand shakes
2) For B's: 3*12=36 hand shakes (excluding hand shake with A bcs it ocurred in 1)
3) For C's: 3*9=27 hand shakes (excluding hand shake with A and B bcs it ocurredin 1 and 2)
4) For D's: 3*6=18 hand shakes (excluding hand shake with A, B and C bcs it ocurred in 1, 2 and 3).
5) For E's: 3*3=9 hand shakes

=45+36+27+18+9=135
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Director
Joined: 14 Jan 2007
Posts: 780
Followers: 2

Kudos [?]: 71 [0], given: 0

I agree with X&Y 's explanation.
Manager
Joined: 12 Feb 2007
Posts: 167
Followers: 1

Kudos [?]: 1 [0], given: 0

yes I also believe X&Y is correct, and therefore he may want to double check his answer.
VP
Joined: 06 Feb 2007
Posts: 1023
Followers: 20

Kudos [?]: 131 [0], given: 0

The OA is D indeed.
X&Y, great job!
Director
Joined: 13 Mar 2007
Posts: 545
Schools: MIT Sloan
Followers: 4

Kudos [?]: 28 [0], given: 0

another way to solve it ..

total 18 folks ..

so total handshakes possible = 18c2 = 153

they dont shake hands within the group ..

within the group = 3c2 = 3 handshakes , hence for 6 groups = 18 handshakes

so, req number = 153 - 18 = 135
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