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# p&c based puzzles

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Intern
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p&c based puzzles [#permalink]

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17 Jan 2014, 20:54
in how many ways can we distribute

(1)5 different balls to 3 different baskets

(2)5 identical balls to 3 different baskets

(3)5 different balls to 3 identical baskets

(4)5 identical balls to 3 identical baskets
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TWICKKKK

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Joined: 03 Feb 2013
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Re: p&c based puzzles [#permalink]

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18 Jan 2014, 03:13
vempatichaitu wrote:
in how many ways can we distribute

(1)5 different balls to 3 different baskets

(2)5 identical balls to 3 different baskets

(3)5 different balls to 3 identical baskets

(4)5 identical balls to 3 identical baskets

Lets tackle the easier question first

2) 5 identical balls to 3 different baskets

This falls into the category called Similar -> Different
As it is similar to different, we don't have to do selection as all of them are same. Only arrangement is required.

A + B + C = 5 => 7C2 = 21

(4)5 identical balls to 3 identical baskets

As again 5 identical balls, there is no selection of balls.

S -> S

Lets say 5 balls goes into the 5 1st basket but as baskets are also identical, hence it doesn't matter. - So we can differentiate based on numbers.

5, 0, 0 - 1 way
4, 1, 0 - 1 way
3, 2, 0 - 1 way
3,1,1 - 1 way
2, 2, 1 - 1 way

Hope I have counted all the possibilities - 5 ways.

(3)5 different balls to 3 identical baskets

D -> S

As the balls are different, the selection matters.

5, 0, 0 - can be only 1 way as all the balls are in a basket and baskets are all identical.
4, 1, 0 - 5C4 - 5 ways.
3, 2, 0 - 5C3 - 15 ways
3, 1, 1 - 5C3 * 2C1 = 15 * 2 = 30 ways.
2,2,1 - 5C2 * 3C2 = 15 * 3 = 45 ways.

Again I hope I have covered all the possibilities.

So total number of ways 96 ways.

(1)5 different balls to 3 different baskets

Now this is a biggie. Selection and arrangement both are required.

D-> D

5, 0, 0 -> 5C5 (Selection) * 3C1 (Arrangement) = 3 ways
4, 1, 0 -> 5C4 * 3C1 * 1C1 * 2C1 = 5 * 3 * 2 = 30 ways
3, 2, 0 -> 5C3 * 3C1 * 3C2 * 2C1 = 15 * 3 * 2 = 90 ways
3, 1, 1 -> 5C3 * 3C1 * 2C1 * 2C1 * 1C1 * 1C1 = 15 * 3 * 2 * 2 = 180 ways
2, 2, 1 -> 5C2 * 3C1 * 3C2 * 2C1 * 1C1 * 1C1 = 15 * 3 * 3 * 2 = 270 ways

Hope I have covered all the possibilities again.

573 ways.

Not sure If I have covered all the possibilities. I can bet on you that apart from 1 st one, other cases may not come in GMAT in the near future. This I learnt from CAT prep days.
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Re: p&c based puzzles [#permalink]

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18 Jan 2014, 05:48
kinjiGC wrote:
vempatichaitu wrote:
in how many ways can we distribute

(1)5 different balls to 3 different baskets

(2)5 identical balls to 3 different baskets

(3)5 different balls to 3 identical baskets

(4)5 identical balls to 3 identical baskets

Lets tackle the easier question first

2) 5 identical balls to 3 different baskets

This falls into the category called Similar -> Different
As it is similar to different, we don't have to do selection as all of them are same. Only arrangement is required.

A + B + C = 5 => 7C2 = 21

(4)5 identical balls to 3 identical baskets

As again 5 identical balls, there is no selection of balls.

S -> S

Lets say 5 balls goes into the 5 1st basket but as baskets are also identical, hence it doesn't matter. - So we can differentiate based on numbers.

5, 0, 0 - 1 way
4, 1, 0 - 1 way
3, 2, 0 - 1 way
3,1,1 - 1 way
2, 2, 1 - 1 way

Hope I have counted all the possibilities - 5 ways.

(3)5 different balls to 3 identical baskets

D -> S

As the balls are different, the selection matters.

5, 0, 0 - can be only 1 way as all the balls are in a basket and baskets are all identical.
4, 1, 0 - 5C4 - 5 ways.
3, 2, 0 - 5C3 - 15 ways
3, 1, 1 - 5C3 * 2C1 = 15 * 2 = 30 ways.
2,2,1 - 5C2 * 3C2 = 15 * 3 = 45 ways.

Again I hope I have covered all the possibilities.

So total number of ways 96 ways.

(1)5 different balls to 3 different baskets

Now this is a biggie. Selection and arrangement both are required.

D-> D

5, 0, 0 -> 5C5 (Selection) * 3C1 (Arrangement) = 3 ways
4, 1, 0 -> 5C4 * 3C1 * 1C1 * 2C1 = 5 * 3 * 2 = 30 ways
3, 2, 0 -> 5C3 * 3C1 * 3C2 * 2C1 = 15 * 3 * 2 = 90 ways
3, 1, 1 -> 5C3 * 3C1 * 2C1 * 2C1 * 1C1 * 1C1 = 15 * 3 * 2 * 2 = 180 ways
2, 2, 1 -> 5C2 * 3C1 * 3C2 * 2C1 * 1C1 * 1C1 = 15 * 3 * 3 * 2 = 270 ways

Hope I have covered all the possibilities again.

573 ways.

Not sure If I have covered all the possibilities. I can bet on you that apart from 1 st one, other cases may not come in GMAT in the near future. This I learnt from CAT prep days.

5 different balls to 3 different baskets
the answer is going to be 3^5 ways as each ball has an option to go into any of the 3 baskets

and your answer for the case of- 5 different balls to 3 identical baskets needs to be corrected also

5, 0, 0 - can be only 1 way as all the balls are in a basket and baskets are all identical.
4, 1, 0 - 5C4 - 5 ways.
3, 2, 0 - 5C3 - 15 ways
3, 1, 1 - 5C3 * 2C1 = 15 * 2 = 30 ways /(2!)
2,2,1 - 5C2 * 3C2 = 15 * 3 = 45 ways /(2!)(as these cases have baskets taking the same no.of balls)
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TWICKKKK

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Re: p&c based puzzles [#permalink]

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20 Jan 2014, 12:37
If you're looking for distribution of balls into baskets -- a highly advanced topic on the GMAT -- take a look at the Permutation/Combinatoric video overview here:
http://www.gmatpill.com/gmat-practice-t ... ons-lesson

If you jump to the 12:00 minute mark - you'll see a distribution of marbles example. You can see even more examples and variations of combination/permutation questions (password combos, poker probability, creating various subgroups within larger groups, coin tosses, letter arrangements) inside the PS Power Pill series in the gmatpill course.
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Re: p&c based puzzles   [#permalink] 20 Jan 2014, 12:37
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# p&c based puzzles

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