A woman has seven cookies - four chocolate chip and three

Question

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

Bunuel · Accepted Answer

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.

ScottTargetTestPrep · Answer

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

JusTLucK04 · Answer

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

v12345 · Answer

the approach to consider 7th left over cookie for charity is good
it helped me to cut down time for question solving

imslogic · Answer

Dear all:

I tried to do the question by subtracting the violations - Deborah and Kim eat different cookies - from the total arrangements. The total arrangements can be calculated as 7!/(4!3!)=35. In my calculation, each arrangement case for the violation produces 4 outcomes with 4 cases. The first case is Kim and Deborah eat cc and oatmeal respectively with 1 cc left over. The second case is Kim and Deborah eat cc and oatmeal with 1 oatmeal left over. The calculation for the first case assuming order is K,D,N,R,M,T is (4×3×3×2×2×1)/(3×2×1×3×2×1) =4. The denominator accounts for the duplicates. The answer I get is 35-16=19. I'd appreciate if someone can help me figure out where I am going wrong with my calculation.

Posted from my mobile device

KarishmaB · Answer

Yes, it is correct.

5C4 * 4! = 5!
So either we pick 4 cookies out of 5 and distribute those 4 in 4! ways
or
We distribute 5 in 5! ways because there are 4 children and one Mr V (or charity).

If say there were 6 cookies and 4 children, we could pick 4 cookies out of 6 in 6C4 ways and distribute them in 4! ways. This would give us (6*5)/2 * 4! = 6!/2
or
We could distribute to 6 people in 6! ways because we have 4 children and 2 identical Mr Vs. Hence, we would divide 6! by 2 to get 6!/2

ThatDudeKnows · Answer

You're close.

Case 1: 3 Chocolate and 3 Oatmeal
K has 6 options.
No matter what K chooses, D has 3 options.
The third to choose has 4 options.
The fourth to choose has 3 options.
The fifth to choose has 2 options.
The sixth to choose has 1 option.

6*3*4*3*2*1 / 3*2*1*3*2*1 = 3*4 = 12

Case 2A: 4 Chocolate and 2 Oatmeal; K chooses Chocolate and D chooses Oatmeal
K has 4 options.
D has 2 options.
The third to choose has 4 options.
The fourth to choose has 3 options.
The fifth to choose has 2 options.
The sixth to choose has 1 option.

4*2*4*3*2*1 / 4*3*2*1*2*1 = 4

Case 2B: 4 Chocolate and 2 Oatmeal; K chooses Oatmeal and D chooses Chocolate
K has 2 options.
D has 4 options.
The third to choose has 4 options.
The fourth to choose has 3 options.
The fifth to choose has 2 options.
The sixth to choose has 1 option.

Same as 2A = 4

Total disallowed options is 12+4+4 = 20.

30-20 = 15

Answer choice D.

Sans8 · Answer

Please tell why my reasoning is wrong?We have 4 Chocolate & 3 Oatmeal cookie
  Case: 1  (Kim and Deborah eats oatmeal cookie)
We're left with- 4Choc& 1 Oatmeal cookie
Now,
>>All 4 chocolate cookies are distributed: 4C4 = 1 way
>> 3 Choc and 1 oatmeal is given: 4C3*1C1*(4!/3!)=16 ways

Case: 2  (Kim and Deborah eats Chocolate cookie)
We're left with- 2Choc& 3Oatmeal cookie
>>3 Oatmeal and 1 chocolate cookie: 3C3*2C1*(4!/3!)=8ways
>>2 Oatmeal and 2 chocolate cookie: 2C2*3C2*(4!/2! 2!)= 18 ways

Total= 43

Where am i going wrong in my though process??
 chetan2u   Bunuel   bb   KarishmaB

chetan2u · Answer

You are arranging these cookies which is wrong and also the calculations are wrong. 
See the red portion.

Case: 1  (Kim and Deborah eats oatmeal cookie)
We're left with- 4Choc& 1 Oatmeal cookie
Now,
>>All 4 chocolate cookies are distributed: 4C4 = 1 way
>>  Choose 3 friends and give them chocolate, fourth will automatically get oatmeal 3 Choc and 1 oatmeal is given: 4C3*1C1* (4!/3!)=16 =4 ways

Case: 2  (Kim and Deborah eats Chocolate cookie)
We're left with- 2Choc& 3Oatmeal cookie
>>  Choose 3 friends and give them oatmeal, fourth will automatically get choc  => 4C3*1C1=4
3 Oatmeal and 1 chocolate cookie: 4C3*2C1 *(4!/3!)=8 =2 ways

>>  Choose 2 friends and give them chocolate, other two will automatically get oatmeal  => 4C2*2C2=6
2 Oatmeal and 2 chocolate cookie: 2C2*3C2* (4!/2! 2!)= 18 ways =6 ways

Total=  43  1+4+4+6=15

KarishmaB · Answer

The highlighted are incorrect.

>> 3 Choc and 1 oatmeal is given: 4C3*1C1*(4!/3!)=16 ways  
You cannot use 4C3 to select 3 chocolate cookies out of 4. Note that all chocolate cookies are the same. So we have only 1 way of selecting 3 chocolate cookies out of 4 - pick any 3.

>>3 Oatmeal and 1 chocolate cookie: 3C3*2C1*(4!/3!)=8ways
>>2 Oatmeal and 2 chocolate cookie: 2C2*3C2*(4!/2! 2!)= 18 ways

Similarly, here you cannot use 2C1 to select 1 chocolate cookie out of 2 because both are same. Again, you cannot use 3C2 because all chocolate cookies are the same. Solve without these figures and you will get your answer.

kriti126 · Answer

Can anyone tell me why we're are not multiplying the cases with the 2 cookies chosen for Deborah and Kim? 
If they are given chocochip cooking then 4P2×(5P4÷3!2!) ?
For selecting and arranging those 2 selected cookies.

Bunuel · Answer

Because each chocolate chip cookie and each oatmeal cookie is identical, there’s no need to account for different arrangements of the selected cookies.

bumpbot · Answer

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A woman has seven cookies - four chocolate chip and three

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