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Three friends are playing a game in which each person

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Three friends are playing a game in which each person [#permalink] New post 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?"

Please show work.
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 [#permalink] New post 22 Feb 2005, 10:34
9 ways:

3 people x 3 ways...

or:

call each item the following:

C = clenched fist
O = open palm
T = two extended fingers

here are the possible combinations:

CCC
COT
CTO
TTT
TCO
TOC
OOO
OTC
OCT
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 [#permalink] New post 22 Feb 2005, 10:39
There is also COO. But I'd like to know how to do it with formulas, not writing each one out. [/quote]
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 [#permalink] New post 22 Feb 2005, 11:18
First person signs* second person signs*third persone signs
=3*3*3=27
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 [#permalink] New post 22 Feb 2005, 11:27
The correct answer is 10 - how can I get this using a formula?
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 [#permalink] New post 22 Feb 2005, 12:07
I fell for the MRQ trap(mis-read the apquestion).This needs lil bit thinking.Will do it later.
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 [#permalink] New post 23 Feb 2005, 10:28
math masters are silent...
bad sign :P
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 [#permalink] New post 23 Feb 2005, 10:43
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.
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 [#permalink] New post 23 Feb 2005, 10:45
hornsfan2005 wrote:

here are the possible combinations:

CCC
COT
CTO
TTT
TCO
TOC
OOO
OTC
OCT


Not right. COT and CTO are considered the same in this question. It would be:
COT
CCC
OOO
TTT
CCO
CCT
OOC
OOT
TTO
TTC
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 [#permalink] New post 23 Feb 2005, 11:36
Thanks HongHu. I too first thought it to be 3*3*3 , but when I saw the answer by LRod, I worked it out. A lil tricky qn.
Good explanation,though.
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 [#permalink] New post 25 Feb 2005, 04:09
hmm a tricky one. lets say the actions are a,b,c

then if all 3 have same signs.. there is only 3 poss

aaa
bbb
ccc
which is 3 in total


if all have diff signs there is only 1 poss
abc
(remember the oredr doesn't matter here)

if any 2 signs are same and 1 diff then
aab
bba
cca
ccb
aac
bbc
that is 6 in all

thus 1 + 3 + 6 = 1

I was surprised to c 10 as the answer. It was then I changed my thought process.
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 [#permalink] New post 25 Feb 2005, 12:35
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)

Is it confusing?
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 [#permalink] New post 25 Feb 2005, 14:57
anirban16 wrote:
The 1 is to include the term in which all are different (which is not included in the first term)


I don't think this is right?
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 [#permalink] New post 25 Feb 2005, 17:03
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.
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 [#permalink] New post 25 Feb 2005, 22:26
Deleted. I'm obviously hallusinating at the time. :oops:

Last edited by HongHu on 26 Feb 2005, 07:49, edited 1 time in total.
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 [#permalink] New post 25 Feb 2005, 23:29
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)

So, total way swill be 9+1.

Hope I made it clear this time.
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 [#permalink] New post 26 Feb 2005, 07:48
Yes, that's definitely correct. :b:
  [#permalink] 26 Feb 2005, 07:48
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