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Re: Five balls of different colors are to be placed in three [#permalink]
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15 Nov 2013, 19:38
when can we use (n1)C(r1) ? It is equal to number of ways of dividing n things tor people where each can receive atleast 1? isnt it the case here?



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Re: Five balls of different colors are to be placed in three [#permalink]
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15 Nov 2013, 23:52
adityapagadala wrote: when can we use (n1)C(r1) ? It is equal to number of ways of dividing n things tor people where each can receive atleast 1? isnt it the case here? The n things need to be identical to use this formula.
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Re: Five balls of different colors are to be placed in three [#permalink]
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16 Nov 2013, 01:05
oh got it...thanks for pointing out the difference..!!



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Re: Five balls of different colors are to be placed in three [#permalink]
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17 Nov 2013, 07:10
VeritasPrepKarishma wrote: The boxes are different B1, B2, B3 When you split the balls 311, which box gets 3 balls? You first select a box which gets 3 balls. You can do this in 3 ways  select B1 or B2 or B3. Now select 3 of the 5 balls for this box in 5C3 ways. Now you have 2 balls and 2 boxes one for each ball so you can distribute them in 2 ways. This gives 3*5C3*2 = 60.
Same problem for 221 case. You multiply by another 3 and you get 3*30 = 90
Total 60+90 = 150
thank you so much for answering me.



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Re: Five balls of different colors are to be placed in three [#permalink]
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25 Nov 2013, 08:13
For at least 1 ball in each box, 2 kind of arrangements possible: (1) 311 or (2) 221
(1) First choose balls for this arrangement using combinations formula: 5C3 x 2C1 x 1C1; then Permute the arrangement: 3! / 2!; so we get: 5C3 x 2C1 x 1C1 x 3! / 2!=60;
(2) Similarly: 5C2 x 3C2 x 1C1 x 3! / 2!=90;
do (1)+(2)=60+90=150



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Re: Five balls of different colors are to be placed in three [#permalink]
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12 Dec 2013, 23:04
We have to ways of putting ball in the boxes
1) 2+2+1 which gives (5!3/2!2!1!)=90 ways, where 5! is number of ways to arrange 5 balls in a row, multiply by 3 as we have 3 boxes, divide by 2!2!1! to unarrange balls in the boxes as the order doesn't matter 2) 3+1+1 which gives 5!3/3!1!1!=60 ways, the same reasoning as in case 1 3) 90+60=150 adding, as either case works for us



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Re: Five balls of different colors are to be placed in three [#permalink]
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02 Feb 2014, 00:46
Leveraging off of Bunuel's approach, but attempting to tackle from a slightly different angle. 311:Ball Combos (Choose 3 from 5)*(Choose 1 from 2 remaining)*(Choose 1 from 1 remaining) 5C3 * 2C1 * 1C1 = 20 Grouping Arrangements (Choose 2 "1 groupings" from 3 slots with the 3 grouping automatically filling the vacant spot) 3C2 = 3 20*3=60 221:Ball Combos (Choose 2 from 5)*(Choose 2 from 3 remaining)*(Choose 1 from 1 remaining) 5C2 * 3C2 *1C1 = 30 Grouping Arrangements (Choose 2 "2 groupings" from 3 slots, with the 1 grouping automatically filling the vacant spot) 3C2 = 3 30*3=90 60+90=150
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Re: Five balls of different colors are to be placed in three [#permalink]
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30 Jun 2016, 08:54
Bunuel wrote: avaneeshvyas wrote: Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60 B. 90 C. 120 D. 150 E. 180
Please provide a small note of explanation for all the combinations used in the solution. We can have the following two distributions: 1. 311 one box gets three balls and the remaining two boxes get one ball each.The number of ways to distribute the balls for this case is \(3*C^3_5*2=60\), where 3 is the number of ways to choose which box gets 3 balls (we have 3 boxes, thus 3 choices for that), \(C^3_5\) is the number of ways to choose which 3 balls out of 5 will go to that box, and 2 is the number of ways to distribute the remaining 2 balls in the remaining two boxes. [b]2. 122 one box gets one ball and the remaining two boxes get two balls each.The number of ways to distribute the balls for this case is \(3*5*C^2_4=90\), where 3 is the number of ways to choose which box gets 1 balls (we have 3 boxes, thus 3 choices for that), 5 is the number of ways to choose which ball out of 5 will go to that box, and \(C^2_4\) is the number of ways to chose which 2 balls out of 4 balls left will go to the second box (the remaining 2 balls will naturally go to the third box).[/b] Total: 60+90=150. Answer: D. Hope it's clear. On stage 2 (122) shouldn't it be 3*5C1*2*4C2 ? My reasoning is that for choosing which of the 3 boxes will have 1 ball we have to multiply by 3. Then 5C1 is for which of the 5 balls will go in that box. Then there are 2 boxes left that will contain 2 balls, hence we multiply by 2. Then the remaining balls could go in 4C2 ways to one of those boxes. What am I doing wrong?



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Re: Five balls of different colors are to be placed in three [#permalink]
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30 Jun 2016, 09:06
Mariwa wrote: Bunuel wrote: avaneeshvyas wrote: Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60 B. 90 C. 120 D. 150 E. 180
Please provide a small note of explanation for all the combinations used in the solution. We can have the following two distributions: 1. 311 one box gets three balls and the remaining two boxes get one ball each.The number of ways to distribute the balls for this case is \(3*C^3_5*2=60\), where 3 is the number of ways to choose which box gets 3 balls (we have 3 boxes, thus 3 choices for that), \(C^3_5\) is the number of ways to choose which 3 balls out of 5 will go to that box, and 2 is the number of ways to distribute the remaining 2 balls in the remaining two boxes. [b]2. 122 one box gets one ball and the remaining two boxes get two balls each.The number of ways to distribute the balls for this case is \(3*5*C^2_4=90\), where 3 is the number of ways to choose which box gets 1 balls (we have 3 boxes, thus 3 choices for that), 5 is the number of ways to choose which ball out of 5 will go to that box, and \(C^2_4\) is the number of ways to chose which 2 balls out of 4 balls left will go to the second box (the remaining 2 balls will naturally go to the third box).[/b] Total: 60+90=150. Answer: D. Hope it's clear. On stage 2 (122) shouldn't it be 3*5C1*2*4C2 ? My reasoning is that for choosing which of the 3 boxes will have 1 ball we have to multiply by 3. Then 5C1 is for which of the 5 balls will go in that box. Then there are 2 boxes left that will contain 2 balls, hence we multiply by 2. Then the remaining balls could go in 4C2 ways to one of those boxes. What am I doing wrong? When you choose with \(C^2_4\) balls for the second box, the remaining 2 automatically go to the third box, so non need to multiply by 2 here.
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