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Five balls of different colors are to be placed in three

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Re: Five balls of different colors are to be placed in three [#permalink] New post 15 Nov 2013, 18:38
when can we use (n-1)C(r-1) ?
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] New post 15 Nov 2013, 22:52
Expert's post
adityapagadala wrote:
when can we use (n-1)C(r-1) ?
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] New post 16 Nov 2013, 00:05
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] New post 17 Nov 2013, 06:10
VeritasPrepKarishma wrote:

The boxes are different B1, B2, B3
When you split the balls 3-1-1, 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 2-2-1 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] New post 25 Nov 2013, 07:13
For at least 1 ball in each box, 2 kind of arrangements possible: (1) 3-1-1 or (2) 2-2-1

(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] New post 12 Dec 2013, 22:04
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 un-arrange 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] New post 01 Feb 2014, 23:46
Expert's post
Leveraging off of Bunuel's approach, but attempting to tackle from a slightly different angle.

3-1-1:

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

2-2-1:

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] 01 Feb 2014, 23:46
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