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A woman has seven cookies - four chocolate chip and three

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A woman has seven cookies - four chocolate chip and three [#permalink]

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A woman has seven cookies - four chocolate chip and three oatmeal. She gives one cookie to each of her six children: Nicole, Ronit, Kim, Deborah, Mark, and Terrance. If Deborah will only eat the kind of cookie that Kim eats, in how many different ways can the cookies be distributed?

(A) 5040
(B) 50
(C) 25
(D) 15
(E) 12
[Reveal] Spoiler: OA

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Last edited by Bunuel on 18 Sep 2012, 05:03, edited 1 time in total.
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Re: Combinations cookie problem [#permalink]

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New post 06 Feb 2011, 14:49
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rxs0005 wrote:
A woman has seven cookies--four chocolate chip and three oatmeal. She gives one cookie to each of her six children: Nicole, Ronit, Kim, Deborah, Mark, and Terrance. If Deborah will only eat the kind of cookie that Kim eats, in how many different ways can the cookies be distributed?

(A) 5040
(B) 50
(C) 25
(D) 15
(E) 12


We have to distribute 7 cookies between 6 kids and give 7th leftover cookie for charity, so distribute 7 cookies between 7 entities: Deborah, Kim, 1, 2, 3, 4, 5.

If Deborah and Kim get chocolate cookies then there will be 2 chocolate and 3 oatmeal cookies to distribute between 5: 5!/(2!3!) = 10 (assign 5 cookies out of which 2 chocolate and 3 oatmeal cookies are identical to 5 people);

If Deborah and Kim get oatmeal cookies then there will be 4 chocolate and 1 oatmeal cookie to distribute between 5: 5!/(4!1!) = 5;

10 + 5 = 15.

Answer: D.
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Re: Combinations cookie problem [#permalink]

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New post 06 Feb 2011, 15:07
Bunuel

thanks for the explanation , my Q is why do we divide 5! by 3! and 2! (because of the repeatsi assume) can you explain that conceptually i know i am missing something fundamental here

my Q is why do we do that or how do we get to 5C2 without that line of reasoning if its there at all...

thanks
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Re: Combinations cookie problem [#permalink]

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New post 06 Feb 2011, 15:10
rxs0005 wrote:
Bunuel

thanks for the explanation , my Q is why do we divide 5! by 3! and 2! (because of the repeatsi assume) can you explain that conceptually i know i am missing something fundamental here

my Q is why do we do that or how do we get to 5C2 without that line of reasoning if its there at all...

thanks


This might help: counting-ps-106762.html
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Re: A woman has seven cookies - four chocolate chip and three [#permalink]

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New post 24 Apr 2014, 00:35
AM I conceptually correct when I say that 5c4 is the number of ways to choose 4 cookies out of 5 (As we have 4 children)...and then the cookies will arrange themselves among children in 4! ways
Of the cookies 3 are identical and other 2 are identical
So---> 5c4*4!/3!*2! = 10

Will this rule apply even when the cookies are 10 and the children are 4...or is it applicable coincidentally for this very specific question
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Re: A woman has seven cookies - four chocolate chip and three [#permalink]

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Bunuel wrote:
rxs0005 wrote:
A woman has seven cookies--four chocolate chip and three oatmeal. She gives one cookie to each of her six children: Nicole, Ronit, Kim, Deborah, Mark, and Terrance. If Deborah will only eat the kind of cookie that Kim eats, in how many different ways can the cookies be distributed?

(A) 5040
(B) 50
(C) 25
(D) 15
(E) 12


We have to distribute 7 cookies between 6 kids and give 7th leftover cookie for charity, so distribute 7 cookies between 7 entities: Deborah, Kim, 1, 2, 3, 4, 5.

If Deborah and Kim get chocolate cookies then there will be 2 chocolate and 3 oatmeal cookies to distribute between 5: 5!/(2!3!) = 10 (assign 5 cookies out of which 2 chocolate and 3 oatmeal cookies are identical to 5 people);

If Deborah and Kim get oatmeal cookies then there will be 4 chocolate and 1 oatmeal cookie to distribute between 5: 5!/(4!1!) = 5;

10 + 5 = 15.

Answer: D.


the approach to consider 7th left over cookie for charity is good
it helped me to cut down time for question solving
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A woman has seven cookies - four chocolate chip and three [#permalink]

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New post 15 Oct 2016, 06:40
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I don't understand the "the 7th is for charity"...

Don't we have precisely to consider whether is 7th is chocolate or oatmeal ? and thus study different cases ?

I have worked that way :

case 1 : K and D choose the chocolate
so, it remains 2C + 30

sub-case a : the 7th is chocolate
so, 1C+30 for four children 4!/(1!3!)

sub-case b : the 7th is oatmeal
so, 2C+2O for four children 4!/(2!2!)

... and so on

What it is wrong with this approach ?

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Re: A woman has seven cookies - four chocolate chip and three [#permalink]

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Re: A woman has seven cookies - four chocolate chip and three   [#permalink] 31 Oct 2017, 22:16
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