Author 
Message 
TAGS:

Hide Tags

Director
Joined: 07 Jun 2004
Posts: 539
Location: PA

A woman has seven cookies  four chocolate chip and three
[#permalink]
Show Tags
Updated on: 18 Sep 2012, 05:03
Question Stats:
50% (02:40) correct 50% (02:53) wrong based on 306 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
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.




Math Expert
Joined: 02 Sep 2009
Posts: 65012

Re: Combinations cookie problem
[#permalink]
Show Tags
06 Feb 2011, 14:49
rxs0005 wrote: A woman has seven cookiesfour 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.
_________________




Director
Joined: 07 Jun 2004
Posts: 539
Location: PA

Re: Combinations cookie problem
[#permalink]
Show Tags
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



Math Expert
Joined: 02 Sep 2009
Posts: 65012

Re: Combinations cookie problem
[#permalink]
Show Tags
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: countingps106762.html
_________________



Retired Moderator
Joined: 17 Sep 2013
Posts: 315
Concentration: Strategy, General Management
WE: Analyst (Consulting)

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



Manager
Joined: 01 Mar 2015
Posts: 108

Re: A woman has seven cookies  four chocolate chip and three
[#permalink]
Show Tags
11 Oct 2015, 07:57
Bunuel wrote: rxs0005 wrote: A woman has seven cookiesfour 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



Intern
Joined: 01 Jul 2018
Posts: 5

Re: A woman has seven cookies  four chocolate chip and three
[#permalink]
Show Tags
25 Jul 2018, 07:42
Hi BunuelThe 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 cookiesfour 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.



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 11042
Location: United States (CA)

Re: A woman has seven cookies  four chocolate chip and three
[#permalink]
Show Tags
23 Apr 2019, 19:03
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 NRMT as follows: ccoo (and 2 c’s and 2 o’s can be arranged in 4!/(2!2!) = 24/(2 x 2) = 6 ways) cooo (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 NRMT as follows: cccc (and the 4 c’s can be arranged in only 1 way) ccco (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
_________________
5star rated online GMAT quant self study course
See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews



Intern
Joined: 04 Jun 2018
Posts: 4
Location: India
GPA: 4
WE: Marketing (Consumer Products)

Re: A woman has seven cookies  four chocolate chip and three
[#permalink]
Show Tags
07 Jun 2019, 03:31
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



Intern
Joined: 08 Oct 2019
Posts: 2

Re: A woman has seven cookies  four chocolate chip and three
[#permalink]
Show Tags
01 Jun 2020, 06:27
Bunuel wrote: rxs0005 wrote: A woman has seven cookiesfour 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 = 42 = 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 = 32 = 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. 



Intern
Joined: 22 Apr 2020
Posts: 1

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



Intern
Joined: 04 Apr 2020
Posts: 3

Re: A woman has seven cookies  four chocolate chip and three
[#permalink]
Show Tags
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 BunuelThis would interest me as well! Thank you!




Re: A woman has seven cookies  four chocolate chip and three
[#permalink]
04 Jun 2020, 10:30




