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Source: — Problem Solving |
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by shankar.ashwin » Tue Sep 27, 2011 9:55 am
4 bowls

5 * 4 * 3 * 2 = 120 (1st can take any of the 5 marbles, 2nd can take 4 and so on)

P.S Pls post answer choices in PS questions

kakz wrote:There are four distinct bowls and five distinct marbles. In how many ways can you place the five marbles in the four bowls with no restrictions?
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by user123321 » Tue Sep 27, 2011 10:23 am
since no restrictions, if we consider a single marble it can go into any of the four bowls in 4 ways...
//ly second marble can go into any of the four bowls in 4 ways...

so OA is it 4*4*4*4*4 = 4^5 ??

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by kakz » Tue Sep 27, 2011 10:24 am
U answer is incorrect. Correct answer is [spoiler](4)^5[/spoiler]
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by gmatboost » Tue Sep 27, 2011 11:00 am
5! would be the right answer if we needed to place exactly 1 in each bowl (and leave one marble out). As the question stands, [spoiler]4^5 is correct.[/spoiler]
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by knight247 » Tue Sep 27, 2011 11:34 am
@Greg

Bro, how is this problem any different from the following problem?

Four identical balls are thrown into the air. Three children are playing a game in which each child holds a barrel that is to be used to catch the balls. If at the end of the game all of the balls have been caught, in how many ways could the four balls be distributed among the three children?

Why can't we use the formula (n+r-1)C(r-1) as we would in the above case? Hoping you could resolve that!! Thanks
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by user123321 » Tue Sep 27, 2011 11:58 am
here balls are identical...so it will eliminate from possibilities which will be formed when balls are distinct.

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by knight247 » Tue Sep 27, 2011 12:17 pm
@user123321
Got another silly question for you. Why couldn't this answer be (5)^4 because we have 4 bowls and the 5 balls can be placed in any of them? Is there any funda to identify which is the base and which is the exponent in such type of problems. Hoping u could clarify. Thanks
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by gmatboost » Tue Sep 27, 2011 12:23 pm
For each marble, we must select EXACTLY ONE BOWL. (We have four choices.)

It is NOT the case that for each bowl, we must select EXACTLY ONE MARBLE. Each bowl could have 0 or 1 or 2 or 3 or 4.

The "decision" if you will, is being made at the marble level. "Which bowl should I put this in?" Once that decision is over, we never think about that marble again, and it does not affect the decision about future marbles. Each marble is independent of other marbles.

Each bowl is not independent of others. If we put 2 in the first bowl, that affects the number we can put in other bowls. The decisions are not independent, so we cannot use a single product to determine the answer. You could go through every possible scenario for each bowl and add them up, but that would take forever.
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by jlittleaziz » Tue Sep 27, 2011 12:44 pm
Bump
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by user123321 » Tue Sep 27, 2011 12:49 pm
hmm..you can get answer by yourself..just reduce the size of the question..
1)2 distinct bowls(A,B) , 3 distinct marbles(M1,M2,M3)
possibilities are...
A-M1,M2,M3;B-nothing & viceversa - 2 ways
A-M1;B-M2,M3 - 3 ways(for A if you keep M2 second time and M3 third time)
A-M1,M2;B-M3 - 3 ways(for B if you keep M1 second time and M2 third time)
so total is 2+3+3 = 8 = 2^3 ways

what you are asking for is...
2)2 distinct bowls (A,B), 3 same marbles(M,M,M)
A-M,M,M;B-nothing (& vice versa) - 2 ways
A-M,M;B-M(& vice versa) - 2 ways
so total is 2+2 = 4 ways = (3+2-1)C(2-1) ways

hope this helps.
user123321

knight247 wrote:@user123321
Got another silly question for you. Why couldn't this answer be (5)^4 because we have 4 bowls and the 5 balls can be placed in any of them? Is there any funda to identify which is the base and which is the exponent in such type of problems. Hoping u could clarify. Thanks
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