How many ways can 4 prizes be given away to 3 boys, if each : PS Archive
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# How many ways can 4 prizes be given away to 3 boys, if each

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How many ways can 4 prizes be given away to 3 boys, if each [#permalink]

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07 Oct 2003, 19:35
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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

thanks
praetorian
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08 Oct 2003, 02:12
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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08 Oct 2003, 02:35
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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Joined: 29 Aug 2003
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Location: Detroit, MI
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08 Oct 2003, 12:02
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?
CEO
Joined: 15 Aug 2003
Posts: 3460
Followers: 67

Kudos [?]: 863 [0], given: 781

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08 Oct 2003, 13:31
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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08 Oct 2003, 16:51
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.

The answer should be 3^4 = 81.

Amar.
CEO
Joined: 15 Aug 2003
Posts: 3460
Followers: 67

Kudos [?]: 863 [0], given: 781

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08 Oct 2003, 16:57
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.

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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Joined: 29 Aug 2003
Posts: 47
Location: Detroit, MI
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08 Oct 2003, 17:03
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.

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.
08 Oct 2003, 17:03
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# How many ways can 4 prizes be given away to 3 boys, if each

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