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Sandy has a husband and 2 children. She brings at least 1 me
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Updated on: 07 Jul 2020, 11:54

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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

Indicate all possible values.

A 45

B 90

C 120

D 210

E 720

F 5,040

Originally posted by alamin on 26 Mar 2019, 00:18.

Last edited by theBrahmaTiger on 07 Jul 2020, 11:54, edited 2 times in total.

Last edited by theBrahmaTiger on 07 Jul 2020, 11:54, edited 2 times in total.

Edited by Carcass

Re: Sandy has a husband and 2 children. She brings at least 1 me
[#permalink]
05 Jun 2019, 12:08

Any explanations for this please?

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Joined: **03 Dec 2019 **

Posts: **348**

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Re: Sandy has a husband and 2 children. She brings at least 1 me
[#permalink]
02 May 2020, 16:01

4

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.

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.

Re: Sandy has a husband and 2 children. She brings at least 1 me
[#permalink]
08 May 2020, 06:40

1

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.

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

Re: Sandy has a husband and 2 children. She brings at least 1 me
[#permalink]
10 May 2020, 04:44

how to do the math?

Re: Sandy has a husband and 2 children. She brings at least 1 me
[#permalink]
10 May 2020, 07:32

Expert Reply

Exactly the way the mods explained above. It is a combination problem

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

Regards

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

Regards

Retired Moderator

Joined: **03 Dec 2019 **

Posts: **348**

Given Kudos: **0 **

Re: Sandy has a husband and 2 children. She brings at least 1 me
[#permalink]
10 May 2020, 07:59

1

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.

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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Joined: **20 Jun 2019 **

Posts: **181**

Given Kudos: **41 **

Re: Sandy has a husband and 2 children. She brings at least 1 me
[#permalink]
07 Jul 2020, 01:29

1

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.

Request moderators to edit this.

Re: Sandy has a husband and 2 children. She brings at least 1 me
[#permalink]
07 Jul 2020, 11:29

Expert Reply

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

Regards

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

Regards

Retired Moderator

Joined: **10 Apr 2015 **

Posts: **6218**

Given Kudos: **136 **

Re: Sandy has a husband and 2 children. She brings at least 1 me
[#permalink]
07 Jul 2020, 12:28

2

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:**

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.

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

Indicate all possible values.

A 45

B 90

C 120

D 210

E 720

F 5,040

Re: Sandy has a husband and 2 children. She brings at least 1 me
[#permalink]
07 Jul 2020, 12:49

Expert Reply

Thank you Sir you came to rescue us

Re: Sandy has a husband and 2 children. She brings at least 1 me
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22 Sep 2023, 23:20

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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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