Last visit was: 16 Nov 2024, 07:11 It is currently 16 Nov 2024, 07:11

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Verbal Expert
Joined: 18 Apr 2015
Posts: 29962
Own Kudos [?]: 36246 [4]
Given Kudos: 25911
Send PM
Most Helpful Community Reply
avatar
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2273 [13]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM
General Discussion
avatar
Intern
Intern
Joined: 21 Nov 2018
Posts: 21
Own Kudos [?]: 11 [0]
Given Kudos: 0
Send PM
avatar
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 470 [0]
Given Kudos: 0
Send PM
Re: Jane must select three different items for each dinner she [#permalink]
1
Expert Reply
kunalkmr62 wrote:
I used the following approach. Can someone tell what is wrong with this approach -:

1 veg should always be there = 5 ways in which this can be selected
now we are left with 4 veg and 4 non-veg dishes
so no of ways in which 2 items can be chosen from a total of 8 = 8C2= 28
hence total no of ways = 28*5 = 140



you are going wrong because there are repetitions in it..
let me give you an example with a smaller set of numbers...

say 2 veg A and B AND 1 non veg C..
so atleast 1 veg should be there and we are looking for ways to choose two ..
By your method..
choose one of two say 2C1 and then 1 from remaining veg and 1 non veg so 1 out of two therefore 2C1
total 2C1*2C1=2*2=4
But is it so...NO
let us see the 4 ways ..
1) choose A, the other could be B or C.. AB and AC
2) choose B, the other could be A or C.. BA and BC
Isnt AB and BA same, so repetitions...

actual answer would be
both veg - 1
one veg and one non veg - 2C1*1=2
total 3

Hope it helps
avatar
Intern
Intern
Joined: 21 Nov 2018
Posts: 21
Own Kudos [?]: 11 [0]
Given Kudos: 0
Send PM
Re: Jane must select three different items for each dinner she [#permalink]
By my method , to chose 1 veg we have 2 ways and now we need to chose 2 items from the remaining 2 , which can be done in only 1 way or 2C2 ways.
Hence total no of dinner possible = 2 . Same as yours

Posted from my mobile device Image
avatar
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 470 [0]
Given Kudos: 0
Send PM
Re: Jane must select three different items for each dinner she [#permalink]
Expert Reply
kunalkmr62 wrote:
By my method , to chose 1 veg we have 2 ways and now we need to chose 2 items from the remaining 2 , which can be done in only 1 way or 2C2 ways.
Hence total no of dinner possible = 2 . Same as yours

Posted from my mobile device Image


No our answers are different, my answer here would be to choose two out of 3 =3C2=3..

Anyways the reason to tell you about that was the flaw in the approach and why you were getting wrong answers.
If you have got it, it is good.
User avatar
Intern
Intern
Joined: 18 Feb 2019
Posts: 7
Own Kudos [?]: 1 [0]
Given Kudos: 0
Send PM
Re: Jane must select three different items for each dinner she [#permalink]
1
I guess from a total of 8 = 8C2= 28 and hence total no of ways = 28*5 = 140. If you are looking for the top hot air fryer, then https://bestairfryer.reviews/gourmia/ is your choice and you have landed on the right page. It is always wise to compare the products and read the full verdict of a particular appliance. This can help you to make an unbiased decision.
avatar
Intern
Intern
Joined: 01 Mar 2020
Posts: 6
Own Kudos [?]: 3 [0]
Given Kudos: 0
Send PM
Re: Jane must select three different items for each dinner she [#permalink]
1
Hi, my approach to this question was a little different. Can you help me identify where I went wrong please?

Since there are three dishes Jane has to pick of which one has to be vegetarian I used the slot method.

The first combination is veggie, meat, meat. There are 5 veggie dishes she can pick from so the first slot is 5. Then there are 4 meat dishes she can pick from so the second slot is 4. Lastly, there are only 3 meat dishes left that she can pick the second dish from so the third slot is 3. This gives a total of 60.

__5__ * __4__ * __3__

Then the other combination is veggie, veggie, meat which gives you a total of 80.

__5__ * __4__ * __4__

The third combination is veggie, veggie, veggie which gives you a total of 60.

__5__ * __4__ * __3__

60+80+60 = 200 which isn't even an option. Can someone tell me where I went wrong?

Thank you!
avatar
Intern
Intern
Joined: 15 May 2020
Posts: 4
Own Kudos [?]: 0 [0]
Given Kudos: 0
Send PM
Re: Jane must select three different items for each dinner she [#permalink]
Based on the ratio of the dishes, I would choose option b
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5014
Own Kudos [?]: 74 [0]
Given Kudos: 0
Send PM
Re: Jane must select three different items for each dinner she [#permalink]
Hello from the GRE Prep Club BumpBot!

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
Re: Jane must select three different items for each dinner she [#permalink]
Moderators:
GRE Instructor
78 posts
GRE Forum Moderator
37 posts
Moderator
1111 posts
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne