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
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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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