Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 27 Apr 2015, 02:35

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 701 [0], given: 781

How many ways can 4 prizes be given away to 3 boys, if each [#permalink]  07 Oct 2003, 19:35
00:00

Difficulty:

(N/A)

Question Stats:

100% (01:03) 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

thanks
praetorian
CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 701 [0], given: 781

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
SVP
Joined: 03 Feb 2003
Posts: 1609
Followers: 6

Kudos [?]: 75 [0], given: 0

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
Intern
Joined: 29 Aug 2003
Posts: 49
Location: Detroit, MI
Followers: 0

Kudos [?]: 2 [0], given: 0

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: 3469
Followers: 61

Kudos [?]: 701 [0], given: 781

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
Intern
Joined: 29 Aug 2003
Posts: 49
Location: Detroit, MI
Followers: 0

Kudos [?]: 2 [0], given: 0

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: 3469
Followers: 61

Kudos [?]: 701 [0], given: 781

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
Intern
Joined: 29 Aug 2003
Posts: 49
Location: Detroit, MI
Followers: 0

Kudos [?]: 2 [0], given: 0

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.
Similar topics Replies Last post
Similar
Topics:
1 In how many ways 4 prizes be given away to 8 boys? 13 26 Aug 2013, 12:30
5 In how many ways can 5 boys and 3 girls be seated on 8 5 24 Sep 2012, 12:02
1 In how many ways can 4 2 27 Oct 2010, 03:05
How many ways are there to split 8 boys into 4 groups of 3 1 01 Nov 2007, 06:46
How many ways 3 girls and 4 boys can be seated so that girls 5 16 May 2005, 11:18
Display posts from previous: Sort by