Four boys picked up 30 mangoes .In how many ways can they divide them if all mangoes be identical?
please help me to solve it............
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Rahul@gurome
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1 mango can be picked by the 4 boys in 4 ways.likithae wrote:Four boys picked up 30 mangoes .In how many ways can they divide them if all mangoes be identical?
please help me to solve it............
So, 30 mangoes can be picked by the 4 boys in [spoiler]4^30[/spoiler] ways
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4^30 would be the answer if all the mangoes were different. Here, however, the mangoes are all identical. If we give the first person 29 mangoes and the second person 1 mango, we don't care *which* mango the second person gets, because all the mangoes are the same.
This is what is known as a 'partition' problem in counting. I've never seen a partition problem in any real GMAT materials - OG, GMATPrep, GMATFocus or on the real test - so you almost certainly don't need to worry about questions like this. And in the very unlikely event that you did see one, the numbers would be much smaller than in this question, which would allow you to use a few different methods to get to an answer. I explained one method that can be used for similar problems here (scroll well down - but the thread as a whole is a bit entertaining), so I won't go through it again:
www.beatthegmat.com/combination-t41362.html
and using that method, you either get the answer 29C3 = (29*28*27)/3! (if each boy needs to be given at least one mango) or 33C3 = (33*32*31)/3! (if you're allowed to give some of the boys zero mangoes - see Stuart's excellent discussion of this type of problem at the very end of the thread linked to above).
Still, it's *so* unlikely that you'll need any of this on your GMAT that I would *only* recommend understanding the concepts here to a test taker who feels completely confident about all of the standard counting/probability question types, and who is looking for some concept-bending questions to think about. Most test takers should just ignore this altogether.
This is what is known as a 'partition' problem in counting. I've never seen a partition problem in any real GMAT materials - OG, GMATPrep, GMATFocus or on the real test - so you almost certainly don't need to worry about questions like this. And in the very unlikely event that you did see one, the numbers would be much smaller than in this question, which would allow you to use a few different methods to get to an answer. I explained one method that can be used for similar problems here (scroll well down - but the thread as a whole is a bit entertaining), so I won't go through it again:
www.beatthegmat.com/combination-t41362.html
and using that method, you either get the answer 29C3 = (29*28*27)/3! (if each boy needs to be given at least one mango) or 33C3 = (33*32*31)/3! (if you're allowed to give some of the boys zero mangoes - see Stuart's excellent discussion of this type of problem at the very end of the thread linked to above).
Still, it's *so* unlikely that you'll need any of this on your GMAT that I would *only* recommend understanding the concepts here to a test taker who feels completely confident about all of the standard counting/probability question types, and who is looking for some concept-bending questions to think about. Most test takers should just ignore this altogether.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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thank you very much.............OA is 33C3 can u explain me little bit clear........Ian Stewart wrote:4^30 would be the answer if all the mangoes were different. Here, however, the mangoes are all identical. If we give the first person 29 mangoes and the second person 1 mango, we don't care *which* mango the second person gets, because all the mangoes are the same.
This is what is known as a 'partition' problem in counting. I've never seen a partition problem in any real GMAT materials - OG, GMATPrep, GMATFocus or on the real test - so you almost certainly don't need to worry about questions like this. And in the very unlikely event that you did see one, the numbers would be much smaller than in this question, which would allow you to use a few different methods to get to an answer. I explained one method that can be used for similar problems here (scroll well down - but the thread as a whole is a bit entertaining), so I won't go through it again:
www.beatthegmat.com/combination-t41362.html
and using that method, you either get the answer 29C3 = (29*28*27)/3! (if each boy needs to be given at least one mango) or 33C3 = (33*32*31)/3! (if you're allowed to give some of the boys zero mangoes - see Stuart's excellent discussion of this type of problem at the very end of the thread linked to above).
Still, it's *so* unlikely that you'll need any of this on your GMAT that I would *only* recommend understanding the concepts here to a test taker who feels completely confident about all of the standard counting/probability question types, and who is looking for some concept-bending questions to think about. Most test takers should just ignore this altogether.
thank u ........but please explain me clearly to solve these problem..........Ian Stewart wrote:4^30 would be the answer if all the mangoes were different. Here, however, the mangoes are all identical. If we give the first person 29 mangoes and the second person 1 mango, we don't care *which* mango the second person gets, because all the mangoes are the same.
This is what is known as a 'partition' problem in counting. I've never seen a partition problem in any real GMAT materials - OG, GMATPrep, GMATFocus or on the real test - so you almost certainly don't need to worry about questions like this. And in the very unlikely event that you did see one, the numbers would be much smaller than in this question, which would allow you to use a few different methods to get to an answer. I explained one method that can be used for similar problems here (scroll well down - but the thread as a whole is a bit entertaining), so I won't go through it again:
www.beatthegmat.com/combination-t41362.html
and using that method, you either get the answer 29C3 = (29*28*27)/3! (if each boy needs to be given at least one mango) or 33C3 = (33*32*31)/3! (if you're allowed to give some of the boys zero mangoes - see Stuart's excellent discussion of this type of problem at the very end of the thread linked to above).
Still, it's *so* unlikely that you'll need any of this on your GMAT that I would *only* recommend understanding the concepts here to a test taker who feels completely confident about all of the standard counting/probability question types, and who is looking for some concept-bending questions to think about. Most test takers should just ignore this altogether.
