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Three friends are playing a game in which each person [#permalink]
22 Feb 2005, 10:29

"Three friends are playing a game in which each person simultaneously displays one of three hand signs: a clenched fist, an open palm, or two extended fingers. How many different combinations of the signs are possible?"

All of them could be the same sign: C(3,1)
Or they could be all different: C(3,3)
Or it could be any two from the three: C(3,2). However, since there are three people, if there are two signs that must mean one sign is showing by two people. Outcome is C(2,1).

Total outcome = C(3,1)+C(3,3)+C(3,2)*C(2,1)=10

I have to admit I first thought 3*3*3 too. This tells us how important reading the question is.

I liked Vijo's solution. Formulae don't work all the time, just use intuition. But it is very important to distinguish between permutation and combination.

But my simple formula will be

3*3*3/3 + 1

The first term is set of all possible outcomes (permutation)/(dividing by 3 to convert the permutation into combination).

The 1 is to include the term in which all are different (which is not included in the first term)

no of ways = C(atleast two of them are same)+ c(all differrent).
C(atleast two of them are same)
= First 2 same selections can only be aa,bb,cc and 3rd selection can be a,b,c
=3*3=9

c(all differrent)=1

So,total=9+1=10

Again,it is tough to get these ideas in the 2 min time we have in the gmat.It took about 15 mins to get this idea.I would rather go with the method suggested by HongHu, not that complicated.

I am not sure what you mean by filling middle positions.

let me explain what I mean in the method.

3 signs can be anything.All 3 can be same(like aaa,bbb...), or two of same and the other differrent(aab,bbc...),or all differrent(abc) etc.,

One simple way to put it is:
Ways for having atleast two of same signs and any third sign +
ways for having all three differrent signs.

In the first portion you will first pick two fo same signs and the third one can be any sign.This combinations represent all possible ways of having signs except one choice where none of the signs are repeated.In other words, the only choice that is missed out from it is,having all three differrent color i.e the choice of picking abc.

So, for the first porion calculation will be: 3*3 (first 3 represents aa,bb,cc and the second 3 represents a,b,c)

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