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In how many ways can all the 7 sweets be distributed to [#permalink]
06 Jan 2005, 09:50

4

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Difficulty:

(N/A)

Question Stats:

72% (01:43) correct
28% (00:31) wrong based on 58 sessions

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

Re: PS- Permutation/Combination - Sweet [#permalink]
07 Jan 2005, 19:02

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

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 ?

Re: PS- Permutation/Combination - Sweet [#permalink]
02 Oct 2009, 13:44

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. _________________

Re: PS- Permutation/Combination - Sweet [#permalink]
04 Oct 2009, 11:05

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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