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Paula and Anna decided to order a pizza with 6 slices all w
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EgmatQuantExpert
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Paula and Anna decided to order a pizza with 6 slices all w [#permalink] New post  28 Mar 2017, 10:23
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Q.

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

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Saquib
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Paula and Anna decided to order a pizza with 6 slices all w [#permalink] New post  Updated on: 31 Mar 2017, 02:29
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The official solution has been posted. Looking forward to a healthy discussion..:)
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Originally posted by EgmatQuantExpert on 28 Mar 2017, 10:24.
Last edited by EgmatQuantExpert on 31 Mar 2017, 02:29, edited 1 time in total.
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foled
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Re: Paula and Anna decided to order a pizza with 6 slices all w [#permalink] New post  28 Mar 2017, 11:27
each girl chooses 3 slices from 6 and 1 of 2 cakes. This means: 3C6*2
Answer is C
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Re: Paula and Anna decided to order a pizza with 6 slices all w [#permalink] New post  31 Mar 2017, 00:09
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EgmatQuantExpert wrote:
Q.

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


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
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Re: Paula and Anna decided to order a pizza with 6 slices all w [#permalink] New post  31 Mar 2017, 00:15
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foled wrote:
each girl chooses 3 slices from 6 and 1 of 2 cakes. This means: 3C6*2
Answer is C



Hey Foled,

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



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Re: Paula and Anna decided to order a pizza with 6 slices all w [#permalink] New post  22 Apr 2017, 18:01
EgmatQuantExpert wrote:
Q.

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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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?
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Re: Paula and Anna decided to order a pizza with 6 slices all w [#permalink] New post  22 Apr 2017, 19:05
EgmatQuantExpert wrote:
EgmatQuantExpert wrote:
Q.

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


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
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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.
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Re: Paula and Anna decided to order a pizza with 6 slices all w [#permalink] New post  24 Apr 2017, 21:36
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mrdlee23 wrote:

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.



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
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Re: Paula and Anna decided to order a pizza with 6 slices all w [#permalink] New post  24 Apr 2017, 21:39
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Nunuboy1994 wrote:


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?


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
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Re: Paula and Anna decided to order a pizza with 6 slices all w [#permalink] New post  12 Dec 2018, 09:33
EgmatQuantExpert wrote:
EgmatQuantExpert wrote:
Q.

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


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
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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 )
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Paula and Anna decided to order a pizza with 6 slices all w [#permalink] New post  12 Dec 2018, 22:24
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vanam52923 wrote:

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 )


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.
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Re: Paula and Anna decided to order a pizza with 6 slices all w [#permalink] New post  13 Dec 2018, 00:05
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
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