Paula and Anna decided to order a pizza with 6 slices all w

Question

Paula and Anna decided to order a pizza with 6 slices all with different toppings and 2 pastries – 1 Chocolate and 1 Caramel. In how many different ways can Paula and Anna choose their food if they each eat three pizza slices and 1 pastry?

A. 24
B. 36
C. 40
D. 80
E. 120

Thanks,
Saquib
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Official Answer

C

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EgmatQuantExpert · Answer

The official solution has been posted. Looking forward to a healthy discussion..

foled · Answer

each girl chooses 3 slices from 6 and 1 of 2 cakes. This means: 3C6*2
Answer is C

EgmatQuantExpert · Answer

Solution

Given  :
 • There are  6  slices of pizza.
 o Paula and Anna will have  3  each. 
• There are  2  pastries.
 o Paula and Anna will have  1  each.

Approach  :

• The number of ways in which  3  pizza each can be selected for Paula and Anna  = ^6C_3 * ^3C_3 = \frac{720}{36} = 20 
• The number of ways in which  1  pastry each can be selected for Paula and Anna  = ^2C_1* ^1C_1 = 2 
• As they will have  both the pizza AND the pastry , we need to multiply the above two values to get the total number of ways to choose the food  = 20*2 = 40 .
• Hence the correct answer is    Option C   .

Thanks,
Saquib
Quant Expert
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EgmatQuantExpert · Answer

Hey Foled,

Your answer is absolutely correct. But shouldn't you write  ^3C_6  as  ^6C_3 = \frac{{6!}}{{3!.3!}} ??  ^nC_R = \frac{n!}{{r!.(n-r)!}} ]

Thanks,
Saquib
Quant Expert
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Nunuboy1994 · Answer

One thing that I don't understand about this question is that it asks, I think, what are the possible number of arrangments that can be made for both Anna and Paula? If Paula was choosing 3 of 6 distinct slices and 1 of 2 distinct pastries then there would certainly be 40 possible arrangments. Though what are the number of arrangments considering both Paula and Anna? Wouldn't that be more?

mrdlee23 · Answer

Can you please explain why you would do 6c3 * 6c3 (I understand there are two entities to consider, but the other would get whichever is remaining)? I believe it has the same result as 6c3.

EgmatQuantExpert · Answer

Yes mrdlee23, you are correct that the other will get whatever is remaining and 3C3 is equal to 1. I had written that so that it is clear to everyone that for the other person, there is only 1 way to choose the remaining. Therefore 6C3 * 3C3 = 6C3 only.

Regards, Saquib

EgmatQuantExpert · Answer

Hey,

The question asks us just the number of ways of selecting the slices and pastries and we do not need to consider the arrangement. Also, understand that once say Paula chooses whatever she wants, Anna has to take whatever is left. And we don't need to consider separately the arrangement of Paula and Anna here. The 6C3 *3C3, takes into consideration, all the ways in which Paula and Anna can choose the slices.

Regards,
Saquib
Quant Expert
e-GMAT

vanam52923 · Answer

Why we are not considering cases here:
6c3*2c1(if paula chooses then anna has to eat whtver is left )+6c3*2c1(if anna chooses then paula has to eat whtver is left )

EgmatQuantExpert · Answer

Hi  vanam52923 ,

Those are not two different cases. Both give the same outcomes, since, the second person gets whatever that is left over, after the first person makes her choice. (irrespective of whether the first person is Paula or Anna)

Regards,
Sandeep.

shobhiitgupta · Answer

Ma'am since the question is asking in how many ways can can Paula and Anna share their food and we don't know who gets to chose first(which would tell who gets the left out pieces), shouldn't it be 6C3*2C1*2! ?

Clarify my doubt please.

Thanks

Clarify my doubt please.

Thanks

luisdicampo · Answer

Deconstructing the Question

There are 6 distinct pizza slices and 2 distinct pastries.

Each person gets 3 slices and 1 pastry.

Step-by-step

Choose 3 slices for Paula:
 \binom{6}{3}=20

Anna gets the remaining 3 slices.

Now assign the pastries (Chocolate and Caramel):
 2 	ext{ ways}

Total:
 20 \cdot 2 = 40

Answer: C

Paula and Anna decided to order a pizza with 6 slices all w

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Paula and Anna decided to order a pizza with 6 slices all w

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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