A woman has seven cookies - four chocolate chip and three oatmeal. She gives one cookie to each of her six children

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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


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Distributing 6 cookies can be done in the following ways:
3 chocolate, 3 oatmeal
4 chocolate, 2 oatmeal

Pairing Deborah and Kim (D and K) together, let's examine each way above:

D&K can each have a chocolate cookie, in which case there are 4 choices of children for the remaining chocolate cookie. The remaining children get oatmeal.
4 ways

Same logic holds if D&K get oatmeal cookies instead:
4 ways

Now let's examine the 4 chocolate/2 oatmeal scenario:

D&K get chocolate cookies. Two children can be selected from the remaining 4 to receive the other chocolate cookies as follows;
4!/2!2! = 6 ways. The other 2 children get oatmeal.

D&K get oatmeal cookies. The other children get chocolate. 1 way

Total ways 15,D

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BTGmoderatorDC wrote:
Sat Jul 02, 2022 8:53 pm
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


OA D

Source: Manhattan Prep
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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