BTGmoderatorLU wrote: Source: Manhattan Prep
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
Alternate approach:
Case 1: Deborah and Kim each receive an oatmeal cookie
In this case, 4 chocolate chip cookies and 1 oatmeal cookie remain.
The remaining oatmeal cookie has 5 options: Nicole, Ronit, Mark, Terrance, or no one.
Total ways = 5.
Case 2: Deborah and Kim each receive a chocolate chip cookie
In this case, 3 oatmeal cookies and 2 chocolate chip cookies are left, with the result that 2 or 3 of the remaining children each receive an oatmeal cookie.
From :
The number of ways to choose 2 to receive an oatmeal cookie = 4C2 = (4*3)/(2*1) = 6.
The number of ways to choose 3 to receive an oatmeal cookie = 4C3 = (4*3*2)/(3*2*1) = 4.
Total ways = 6+4 = 10.
Case 1 + Case 2 = 5 + 10 = 15.
The correct answer is
D.
Alternatively, we can simply LIST the number of ways the cookies can be distributed among Nicole, Ronit, Mark, or Terrance in each case.
Case 1: Deborah and Kim each receive an oatmeal cookie
In this case, 4 chocolate chip cookies and 1 oatmeal cookie remain for Nicole, Ronit, Mark, and Terrance.
Options:
CCCC
OCCC
COCC
CCOC
CCCO
Total ways = 5.
Case 2: Deborah and Kim each receive a chocolate chip cookie
In this case, 3 oatmeal cookies and 2 chocolate chip cookies remain for Nicole, Ronit, Mark, and Terrance.
Options:
COOO
OCOO
OOCO
OOOC
OOCC
OCOC
OCCO
COOC
COCO
CCOO
Total ways = 10.
Case 1 + Case 2 = 5 + 10 = 15.
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