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In how many ways can all the 7 sweets be distributed to

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In how many ways can all the 7 sweets be distributed to [#permalink]

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06 Jan 2005, 10:50
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In how many ways can all the 7 sweets be distributed to three friends Jack, Jill, John so that Jill always gets 2 sweets and each of them gets at least one sweet
A) 24
B) 6
C) 4
D) 12
E) 3
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Re: PS- Permutation/Combination - Sweet [#permalink]

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07 Jan 2005, 20:02
4?

I get 8 but then I dont see that in the answers, so maybe I am counting twice.

nocilis wrote:
In how many ways can all the 7 sweets be distributed to three friends Jack, Jill, John so that Jill always gets 2 sweets and each of them gets at least one sweet
A) 24
B) 6
C) 4
D) 12
E) 3
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08 Jan 2005, 13:45
4 ways. There is only one way to give 2 sweets to Jill. And rest of 5 has to be given to John and Jack, which has 4 ways (1,4) (4,1), (2,3), (3,2) = 4

So 1x4 = 4 ways
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08 Jan 2005, 22:43
Is there a mathematical way to solve this problem, for instance what if the number of sweets to be distributed was 15 and there might be more than just 3 members with some conditions on minimum distributions etc. Is there a way to solve it then ?

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09 Jan 2005, 23:02
2-1-4
2-4-1
2-3-2
2-2-3

if jill alwas gets 2 then 4 ways must be the answer. anyone have the OA??
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09 Jan 2005, 23:37
I would love to know the mathematical way to solve it too. Pls help.
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Re: PS- Permutation/Combination - Sweet [#permalink]

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02 Oct 2009, 14:34
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7 sweets: JI JI JA _ _ _ _
4C1 = 4
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Re: PS- Permutation/Combination - Sweet [#permalink]

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02 Oct 2009, 14:44
nocilis wrote:
In how many ways can all the 7 sweets be distributed to three friends Jack, Jill, John so that Jill always gets 2 sweets and each of them gets at least one sweet
A) 24
B) 6
C) 4
D) 12
E) 3

I doubt the question sentence framing. Jill always gets 2 sweets and "each of them gets at least one". Each of them...gets...at least...one!!!! Is Jill being counted again? Jill is given off 2 sweets and isn't she happy ever? Does she demand at least one sweet of the remaining five? Are the five sweets supposed to be distributed to each of them (three??) so that every one gets at least one sweet??

Where was this question even picked? Please don't follow useless sources such as these.
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Re: PS- Permutation/Combination - Sweet [#permalink]

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02 Oct 2009, 15:02
The question is poorly written, but it still is a valid/basic combinatorics problem.

7 sweets. Jill must have 2. Jack and John must have at least 1.
How many combinations are there?
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Re: PS- Permutation/Combination - Sweet [#permalink]

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04 Oct 2009, 12:05
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Essentially, this problem can be rewritten as 5 sweets being distributed among Jack and John, with each of them getting at least 1.

You really only need to focus on one person, since whatever Jack gets, John will get the rest. So for Jack there are 4C1 ways (1 to 4 sweets) his total can be distributed, and the remaining goto John. Therefore 4 total ways.
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Re: PS- Permutation/Combination - Sweet [#permalink]

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14 Oct 2009, 04:12
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There is a very easy mathematical way to solve this question

The way is to see how n sweets can be distributed amongst x people

So we should have n sweets and (x-1) divisions.
And just arrange these (n+x-1) objects
arranging them can be done in (n+x-1)C(n) ways

So in this case we have to distribute 3 sweets among two people - as both of them have already got one each

so n=3 x=2
therefore answer is (3+2-1)C2 = 4C3 ways = 4 ways
These are

OOOS
OOSO
OSOO
SOOO

Where O is the sweet and S is the division
Anything left of S will go Jack and right of it will go to John
I hope that helps.
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Re: PS- Permutation/Combination - Sweet [#permalink]

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14 Oct 2009, 04:48
4 ways...manual approach works better
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Re: PS- Permutation/Combination - Sweet [#permalink]

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14 Oct 2009, 04:51
This will help if u r given 20 sweets and 5-6 people.
It is a more generic solution
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Re: PS- Permutation/Combination - Sweet [#permalink]

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05 May 2011, 03:28
manual technique gives 4.
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Re: PS- Permutation/Combination - Sweet [#permalink]

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05 May 2011, 07:23
2,1,4
2,2,3
2,3,2
2,4,1

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Re: PS- Permutation/Combination - Sweet   [#permalink] 05 May 2011, 07:23
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