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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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New post Updated on: 18 Sep 2012, 05:03
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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

Originally posted by rxs0005 on 06 Feb 2011, 14:37.
Last edited by Bunuel on 18 Sep 2012, 05:03, edited 1 time in total.
Edited the question.
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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: A woman has seven cookies - four chocolate chip and three  [#permalink]

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New post 24 Apr 2014, 00:35
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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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New post 23 Apr 2019, 19:03
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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 can denote each child by the first letter of his or her name.

We have 2 scenarios: 1) D and K both have chocolate chip cookies, and 2) D and K both have oatmeal cookies.

Scenario 1:

After D and K have 2 chocolate chip cookies, the woman has 2 chocolate chip (c) and 3 oatmeal (o) cookies for the remaining 4 children. She can distribute the cookies to N-R-M-T as follows:

c-c-o-o (and 2 c’s and 2 o’s can be arranged in 4!/(2!2!) = 24/(2 x 2) = 6 ways)

c-o-o-o (and 1 c and 3 o’s can be arranged in 4!/3! = 24/6 = 4 ways)

We see that there are 10 ways to distribute the cookies in this scenario.

Scenario 2:

After D and K have 2 oatmeal cookies, the woman has 4 chocolate chip (c) and 1 oatmeal (o) cookies for the remaining 4 children. She can distribute the cookies to N-R-M-T as follows:

c-c-c-c (and the 4 c’s can be arranged in only 1 way)

c-c-c-o (and 3 c’s and 1 o can be arranged in 4!/3! = 24/6 = 4 ways)

We see that there are 5 ways to distribute the cookies in this scenario.

Therefore, the woman has 10 + 5 = 15 ways to distribute the 7 cookies to 6 children.

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

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New post 06 Feb 2011, 15:07
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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
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Re: A woman has seven cookies - four chocolate chip and three  [#permalink]

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New post 11 Oct 2015, 07:57
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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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Re: A woman has seven cookies - four chocolate chip and three  [#permalink]

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New post 25 Jul 2018, 07:42
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Hi Bunuel

The reason you don't multiply each part by 5C2 and 3C2 to choose which 2 cookies to give Deborah and Kim is because cookies identical?

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

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New post 07 Jun 2019, 03:31
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can someone explain me where did i go wrong ?
my approach to this question was:
case 1 : when both D and K have oatmeal cookies -
so, we are left with 1 oatmeal cookie and 4 chocolate cookies. so we can either select all 4 chocolate cookies for the rest 4 children i.e 4c4= 1 way; or 1 oatmeal cookie and 3 chocolate cookies i.e 1c1 (for oatmeal cookies) * 4c3 (for selecting 3 out of 4 chocolate cookies) = 1* 4 = 4 ways.
so total = 1 + 4 = 5 ways

case 2: when both D and K select chocolate cookies -
so we are left with 3 oatmeal cookies and 2 chocolate cookies. so we can either select all 3 oatmeal cookies and 1 chocolate cookie for the rest 4 children i.e 3c3 * 2c1 = 1*2= 2 ways ; or select 2 oatmeal cookies and 2 chocolate cookies = 3c2 (for oatmeal) * 2c2 (for chocolate) = 3*1 = 3 ways
so total= 2+3= 5 ways

hence, overall, we have 5+5 = 10 ways
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Re: A woman has seven cookies - four chocolate chip and three  [#permalink]

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New post 01 Jun 2020, 06:27
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.


Hi,
In you solution above, as far as I can see, the order of distribution of cookies is considered. Could you please enlighten me why the order shall be considered here. I tried to solve the problem with the following approach but ended up with a wrong answer:

Solution
-----------------------------------------------------------------------------------------
CASE 1: Deborah + Kim both gets Chocolate cookies
Remaining chocolate cookies = 4-2 = 2
Oatmeal cookies = 3
Total cookies = 2+3 =5
Total number of ways to distribute 5 cookies among remaining 4 children = C (5,4) = 5 ways

CASE 2:Deborah + Kim both gets Oatmeal cookies
Remaining Oatmeal cookies = 3-2 = 1
Chocolate cookies = 4
Total cookies = 1+4 =5
Total number of ways to distribute 5 cookies among remaining 4 children = C (5,4) = 5 ways

Total number of ways = CASE 1 + CASE 2 = 5+5 = 10.
-------------------------------------------------------------------------------
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Re: A woman has seven cookies - four chocolate chip and three  [#permalink]

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New post 02 Jun 2020, 10:25
Can anyone help me with this?
My reasoning:

The 2 girls get 2 chocolates,the other girls can receive either 1,2 or none chocolates.

-Number of ways to pick 2 out the remaining four, who get chocolate while the others get oatmeal: 4C2,
-Number of ways to pick 1, who gets chocolate, the other 3 get oatmeal: 4C1,
-Other 4 get all oatmeal: 1
_____________________________

The 2 girls get 2 Oatmeals:

-Number of ways to pick 1 out of 4, who gets the remaining oatmeal: 4C1;
-Other 4 get chocolate:1.
_____________________________

Total is 16.
Thanks for the help!
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Re: A woman has seven cookies - four chocolate chip and three  [#permalink]

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New post 04 Jun 2020, 10:30
JusTLucK04 wrote:
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


Bunuel
This would interest me as well! Thank you!
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Re: A woman has seven cookies - four chocolate chip and three   [#permalink] 04 Jun 2020, 10:30

A woman has seven cookies - four chocolate chip and three

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