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mand-y
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ps children coins and his pockets 10 Aug 2006, 18:30
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what about this one ?A CHILD HAS 3 POCKETS AND 4 COINS
In how many ways can he put the coins in his pockets ?

PLZ POST your reasoning ?
i Don 't agree with the OA


thanks



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trivikram
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ps children coins and his pockets 10 Aug 2006, 18:39
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4^3 = 64 ways

1st coin can be put in any of the four pockets in 4 ways
2nd coin can be put in any of the four pockets in 4 ways
3rd coin can be put in any of the four pockets in 4 ways

Total = 4*4*4 = 64
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mand-y
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ps children coins and his pockets 10 Aug 2006, 18:42
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trivikram
4^3 = 64 ways

1st coin can be put in any of the four pockets in 4 ways
2nd coin can be put in any of the four pockets in 4 ways
3rd coin can be put in any of the four pockets in 4 ways

Total = 4*4*4 = 64
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:oops: NOPE SORRY
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ps children coins and his pockets 10 Aug 2006, 18:46
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mand-y
trivikram
4^3 = 64 ways

1st coin can be put in any of the four pockets in 4 ways
2nd coin can be put in any of the four pockets in 4 ways
3rd coin can be put in any of the four pockets in 4 ways

Total = 4*4*4 = 64
:oops: NOPE SORRY
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My bad

3^4 = 81 ways

each coin can we put in 3 ways
likeways 3^4 = 81
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mand-y
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ps children coins and his pockets 10 Aug 2006, 18:50
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trivikram
mand-y
trivikram
4^3 = 64 ways

1st coin can be put in any of the four pockets in 4 ways
2nd coin can be put in any of the four pockets in 4 ways
3rd coin can be put in any of the four pockets in 4 ways

Total = 4*4*4 = 64
:oops: NOPE SORRY

My bad

3^4 = 81 ways

each coin can we put in 3 ways
likeways 3^4 = 81
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YES
so repetition is allowed but why is it permutation or combination
thanks
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ps children coins and his pockets 10 Aug 2006, 23:36
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If there're only k possible outcomes for each object, total possible outcomes for n objects is k^n.

To identify the n and k. Do the repetition test. Thing that is replaced back is k and other one is n.

Here once you have put a coin in a pocket you can not put the same coin in another pocket. But if you have put the a coin in one pocket you can put another coin in the pocket. So the pocket can be be replaced.
so k = 3 and n = 4
Answer =3^4 = 81
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ps children coins and his pockets 12 Aug 2006, 17:58
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Got 81 also, each coin can go into any of the 3 pockets ...

hence 3*3*3*3 which is 3^4 = 81
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ps children coins and his pockets 16 Aug 2006, 09:13
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ps_dahiya
If there're only k possible outcomes for each object, total possible outcomes for n objects is k^n.

To identify the n and k. Do the repetition test. Thing that is replaced back is k and other one is n.

Here once you have put a coin in a pocket you can not put the same coin in another pocket. But if you have put the a coin in one pocket you can put another coin in the pocket. So the pocket can be be replaced.
so k = 3 and n = 4
Answer =3^4 = 81
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Hmm.. What if # of pockets = 1 and coins = 2.. Then 1^2 = 1..

Doesn't make sense.. Answer is 2^1.
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ps children coins and his pockets 16 Aug 2006, 10:24
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sharadGmat
ps_dahiya
If there're only k possible outcomes for each object, total possible outcomes for n objects is k^n.

To identify the n and k. Do the repetition test. Thing that is replaced back is k and other one is n.

Here once you have put a coin in a pocket you can not put the same coin in another pocket. But if you have put the a coin in one pocket you can put another coin in the pocket. So the pocket can be be replaced.
so k = 3 and n = 4
Answer =3^4 = 81

Hmm.. What if # of pockets = 1 and coins = 2.. Then 1^2 = 1..

Doesn't make sense.. Answer is 2^1.
How the answer is 2^1???
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How many ways can we put one of 2 coins into pocket..
i.e. pick one coin of 2...
This is 2C1=2..
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ps_dahiya
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ps children coins and his pockets 16 Aug 2006, 10:36
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sharadGmat
ps_dahiya
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[quote="ps_dahiya"]If there're only k possible outcomes for each object, total possible outcomes for n objects is k^n.

To identify the n and k. Do the repetition test. Thing that is replaced back is k and other one is n.

Here once you have put a coin in a pocket you can not put the same coin in another pocket. But if you have put the a coin in one pocket you can put another coin in the pocket. So the pocket can be be replaced.
so k = 3 and n = 4
Answer =3^4 = 81

Hmm.. What if # of pockets = 1 and coins = 2.. Then 1^2 = 1..

Doesn't make sense.. Answer is 2^1.
How the answer is 2^1???

How many ways can we put one of 2 coins into pocket..
i.e. pick one coin of 2...
This is 2C1=2..[/quote]
OK, it means you will put one coin in the pocket and one on the street :no :no
You have to put all coins not only one or two. If we take your case then infact there are four cases. One when coin 1 is put, one when 2nd coin is put, one when both coins are put and one when none of the coins are put. i.e 2^2 cases. But this is not the case.

Read this post casefully.
https://www.gmatclub.com/phpbb/viewtopic.php?t=14706



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