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CEO
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How many ways can 4 prizes be given away to 3 boys, if each [#permalink]
 07 Oct 2003, 20:35
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all the prizes?
1. 81
2. 12
3. 64
4. 27
5. 48
please explain your choice.
thanks
praetorian
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CEO
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stolyar wrote: 4^3=64
nope
1 prize can be distributed in 3 ways
2nd in 3 ways
3rd in 3 ways
4th in 3 ways
thats 3 ^ 4 = 81
A basic question
why do we multiply, and not add?
thanks
praetorian
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SVP
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where is my mistake?
B1 can get P1, or P2, or P3, or P4=4
and
B2 can get P1, or P2, or P3, or P4=4
and
B3 can get P1, or P2, or P3, or P4=4
OR means +
AND means *
4*4*4
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Manager
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stolyar wrote: where is my mistake?
B1 can get P1, or P2, or P3, or P4=4
and
B2 can get P1, or P2, or P3, or P4=4
and
B3 can get P1, or P2, or P3, or P4=4
OR means +
AND means *
4*4*4
How about the possibility that one or two of the 3 boys does not get any prize at all?
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CEO
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amarsesh wrote: How about the possibility that one or two of the 3 boys does not get any prize at all?
Amar,
Can you explain this?
thanks
praetorian
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praetorian123 wrote: amarsesh wrote: How about the possibility that one or two of the 3 boys does not get any prize at all? Amar, Can you explain this? thanks praetorian What I meant to say was that he was not considering all the possibilities. Stolyar's approach looked as if he was assuming that each boy should get 1 prize at least or something like that. So, I was just listing out the other possibilities too. Sorry for any confusion this might have caused.
Your approach is right.
The answer should be 3^4 = 81.
Amar.
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CEO
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amarsesh wrote:
What I meant to say was that he was not considering all the possibilities. Stolyar's approach looked as if he was assuming that each boy should get 1 prize at least or something like that. So, I was just listing out the other possibilities too. Sorry for any confusion this might have caused.
Your approach is right.
The answer should be 3^4 = 81.
Amar.
amar,
how can we approach these kind of problems... i find them different from the regular combination/permutation problems
i come across lot of them ( eg, 10 letters , 5 post boxes)
what is the counting principle that underlies all of these?
i did not come up with a good explanation for why stolyar is wrong in this case.
could you help?
thanks
praetorian
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praetorian123 wrote: amarsesh wrote:
What I meant to say was that he was not considering all the possibilities. Stolyar's approach looked as if he was assuming that each boy should get 1 prize at least or something like that. So, I was just listing out the other possibilities too. Sorry for any confusion this might have caused.
Your approach is right.
The answer should be 3^4 = 81.
Amar.
amar, how can we approach these kind of problems... i find them different from the regular combination/permutation problems i come across lot of them ( eg, 10 letters , 5 post boxes) what is the counting principle that underlies all of these? i did not come up with a good explanation for why stolyar is wrong in this case. could you help? thanks praetorian I saw the 10 letters, 5 boxes problem on a site and had to think of a proper way to solve it logically for about 10 minutes and came upwith the following logic or approach. Since the questions asks us to find the ways for letters, I went out looking for combinations for each letter. Unfortunately, I do not know of a generic way to solve problems like this (How I wish there were a way to do so, though). This problem you put up here was like the letters one. So, it was easy this time.
Amar.
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