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combination sum - need clarification

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by winnerhere » Tue Aug 25, 2009 9:02 am
I need clarification for two different types of sums

1.Consider 5 similar fruits and find the number of ways they be distributed among 3 boys,where each boy gets atleast one fruit.

2.consider 5 different fruit and find the number of ways they be distributed among 3 boys where each boy gets atleast one fruit.
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by shanrizvi » Tue Aug 25, 2009 11:20 am
Well I don't think it matters if the fruits are similar or different :S
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by winnerhere » Tue Aug 25, 2009 11:19 pm
shanrizvi wrote:Well I don't think it matters if the fruits are similar or different :S
well

2 2 1 - five fruits are splitted like this for three boys

(2 orange) (2 orange) and (1 orange) - if fruits are similar

but

(1 orange 1 apple ) (1 pineapple 1 grape) (1 banana)

is different from

(1 pineapple 1 grape) (1 orange 1 apple ) (1 banana)
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by shanrizvi » Wed Aug 26, 2009 6:21 am
I guess if the order matters then it should be 4C2*5!.
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Re: combination sum - need clarification

by tohellandback » Thu Aug 27, 2009 2:09 am
winnerhere wrote:I need clarification for two different types of sums

1.Consider 5 similar fruits and find the number of ways they be distributed among 3 boys,where each boy gets atleast one fruit.

2.consider 5 different fruit and find the number of ways they be distributed among 3 boys where each boy gets atleast one fruit.
Condition 1: it will be the equation

X+Y+Z=5
and the answer is 4!/(2!*2!)--explanation for this result is discussed in some thread by Ian Stuart. If you don't get this..let me know

2) If the question simply was 5 different fruits to be distributed among 3 boys: then answer would be
5^3
and I really think that this is what you wanted to ask. but anyways..

given the condition that each one of the boys must get atleast 1, it becomes a little difficult.
Cases are:



1 2 2 - 5C2*3C2* 3C2= 90 ways

1 3 1- 5C3 * 3= 30 ways
edit: it should be 2*5C3*3=60 ways

total=150 ways
Last edited by tohellandback on Thu Aug 27, 2009 11:14 pm, edited 2 times in total.
The powers of two are bloody impolite!!
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Re: combination sum - need clarification

by winnerhere » Thu Aug 27, 2009 10:58 pm
tohellandback wrote: Condition 1: it will be the equation

X+Y+Z=5
and the answer is 4!/(2!*2!)--explanation for this result is discussed in some thread by Ian Stuart. If you don't get this..let me know
isnt 4c2 = 6

though ur answer also sums upto 6 :)
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Re: combination sum - need clarification

by tohellandback » Thu Aug 27, 2009 11:05 pm
winnerhere wrote:
tohellandback wrote: Condition 1: it will be the equation

X+Y+Z=5
and the answer is 4!/(2!*2!)--explanation for this result is discussed in some thread by Ian Stuart. If you don't get this..let me know
isnt 4c2 = 6

though ur answer also sums upto 6 :)
well you can always write my above expression as 4C2. and in fact I do that. But try doing that when each of the students can get 0 or more. It will get confusing.
The powers of two are bloody impolite!!
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Re: combination sum - need clarification

by winnerhere » Thu Aug 27, 2009 11:07 pm
tohellandback wrote:
winnerhere wrote: given the condition that each one of the boys must get atleast 1, it becomes a little difficult.
Cases are:



1 2 2 - 5C2*3C2* 3C2= 90 ways

1 3 1- 5C3 * 3= 30 ways
total=120 ways

1 1 3
1 2 2
Nice explanation..Thanks THAB


P.S : the bolded partshould have been 5 * 4c3 * 3 = 60

so total 150 ways
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Re: combination sum - need clarification

by winnerhere » Thu Aug 27, 2009 11:11 pm
tohellandback wrote:
winnerhere wrote:
tohellandback wrote: Condition 1: it will be the equation

X+Y+Z=5
and the answer is 4!/(2!*2!)--explanation for this result is discussed in some thread by Ian Stuart. If you don't get this..let me know
isnt 4c2 = 6

though ur answer also sums upto 6 :)
well you can always write my above expression as 4C2. and in fact I do that. But try doing that when each of the students can get 0 or more. It will get confusing.
Well for that we can use (n+r-1)C(r-1)


anyway...Id like to know how u used that formula 4!/2! 2!
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