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Re: Sandy has a husband and 2 children. She brings at least 1 me [#permalink]
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theBrahmaTiger wrote:
If Sandy brings her husband to the restaurant, then there are two plates that would need to be served.

There are 10 kinds of lunches available and 2 lunches would be chosen.

Hence, by combination -> 10C2 -> 45.

Now, if Sandy brings 1 child along with her husband. Then 3 lunches would have to chosen from 10 choices.

10C3 = 120.

Finally, If Sandy brings 2 kids and her husband -> 10C4 = 210.

Therefore, answer is A,C, and D.


The question is not about combination it is about permutation

Let us say that 3 types of lunch (L1, L2 and L3) are available and 2 people have are ordering

In case if they order different dishes

There are three options for first person and 2 options for the second person. So 6 orders are possible.

3p2 =6 while 3c2 =3 .

So the answers will be

10p2 = 90

10p3 = 720

10p4 = 5040
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Re: Sandy has a husband and 2 children. She brings at least 1 me [#permalink]
how to do the math?
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Re: Sandy has a husband and 2 children. She brings at least 1 me [#permalink]
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Exactly the way the mods explained above. It is a combination problem

here a good start https://gmat.economist.com/gmat-advice/ ... gmat-quant

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Re: Sandy has a husband and 2 children. She brings at least 1 me [#permalink]
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RSQUANT wrote:
theBrahmaTiger wrote:
If Sandy brings her husband to the restaurant, then there are two plates that would need to be served.

There are 10 kinds of lunches available and 2 lunches would be chosen.

Hence, by combination -> 10C2 -> 45.

Now, if Sandy brings 1 child along with her husband. Then 3 lunches would have to chosen from 10 choices.

10C3 = 120.

Finally, If Sandy brings 2 kids and her husband -> 10C4 = 210.

Therefore, answer is A,C, and D.


The question is not about combination it is about permutation

Let us say that 3 types of lunch (L1, L2 and L3) are available and 2 people have are ordering

In case if they order different dishes

There are three options for first person and 2 options for the second person. So 6 orders are possible.

3p2 =6 while 3c2 =3 .

So the answers will be

10p2 = 90

10p3 = 720

10p4 = 5040


I don't think order does matter here. If I buy one pizza and one cheesecake, then it is same as ordering one cheesecake and one pizza. They are making an order together, not individually.
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Re: Sandy has a husband and 2 children. She brings at least 1 me [#permalink]
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This is a badly worded question. What does "She brings at least 1 member of her family to dinner for lunch every day. The none offers 10 lunch specials" even mean?! The none offers? 1 member of her family to dinner for lunch?!

Request moderators to edit this.
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Re: Sandy has a husband and 2 children. She brings at least 1 me [#permalink]
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I think is better to move on from this question or questions like this.

if we edit it without having in hand the question as it was originally maybe we could create even more confusion.

will see what GreenLight could say regarding this one

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Re: Sandy has a husband and 2 children. She brings at least 1 me [#permalink]
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It's a bad question for a two reasons.

First, we aren't told whether Sandy eats at the diner. All we know is that she brings 1, 2, or three people to the diner. This might seem like small potatoes, but the test-makers would never allow such ambiguity in a question.

Second (and more importantly), if answer choice A (45) is one of the official answers, then this means Sandy having a hamburger and her husband having kale is the SAME as Sandy having kale and her husband having a hamburger. Those two outcomes seem very different to me.



Quote:
Sandy has a husband and 2 children. She brings at least 1 member of her family to a diner for lunch every day. The diner offers 10 lunch specials. If no one orders the same thing, how many different orders can Sandy`s family make For lunch?

Indicate all possible values.

A 45
B 90
C 120
D 210
E 720
F 5,040
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Re: Sandy has a husband and 2 children. She brings at least 1 me [#permalink]
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Thank you Sir you came to rescue us :)
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Re: Sandy has a husband and 2 children. She brings at least 1 me [#permalink]
Hello from the GRE Prep Club BumpBot!

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