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 GMAT 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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: In how many ways can five girls stand in line if Maggie and Lisa canno [#permalink]

Show Tags

05 Feb 2010, 13:31

1

This post received KUDOS

Expert's post

1. the total number of permutations: 5! = 120 2. Let's consider Maggie and Lisa as one object, then the total number of permutations with Maggie and Lisa together: 4! = 24. 3. Take into account that [Maggie, Lisa] and [Lisa, Maggie] are different. 4. Maggie and Lisa cannot stand next to each other in: 120 - 2*24 = 72 ways. _________________

Re: In how many ways can five girls stand in line if Maggie and Lisa canno [#permalink]

Show Tags

27 Jul 2015, 00:09

can we solve this with \(P^n_r\) formula? _________________

Kudos = Thanks

"Yeah, you can get a nickel for boosting Starfall, but jacking Heal's a ten-day stint in county. Now lifting Faerie Fire, they just let you go for that — it's not even worth the paperwork. But Reincarnation, man! That'll get you life!"

Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!

Re: In how many ways can five girls stand in line if Maggie and Lisa canno [#permalink]

Show Tags

27 Jul 2015, 05:42

GMATinsight wrote:

Subanta wrote:

can we solve this with \(P^n_r\) formula?

Jut a personal Suggestion: It's best to see the steps and work in steps in P&C problems rather than being dependent on formulas

However, Answer to your query is as follows

Quote:

In how many ways can five girls stand in line if Maggie and Lisa cannot stand next to each other?

(A) 112 (B) 96 (C) 84 (D) 72 (E) 60

5 girls can stand in line in \(P^5_5 = 120 ways\)

5 girls can stand in line in such that Maggie and Lisa stand together in \(P^4_4 * P^2_2= 48 ways\)

Total favourable ways of arrangement of five girls such that Maggie and Lisa cannot stand next to each other = 120 - 48 = 72 ways

Answer: option D

Thanks. I usually solve my problems without the formulae, but I have come across many problems where it is easier to use the formulae. I'm only trying to get familiar with the usage of the formulae. _________________

Kudos = Thanks

"Yeah, you can get a nickel for boosting Starfall, but jacking Heal's a ten-day stint in county. Now lifting Faerie Fire, they just let you go for that — it's not even worth the paperwork. But Reincarnation, man! That'll get you life!"

Re: In how many ways can five girls stand in line if Maggie and Lisa canno [#permalink]

Show Tags

27 Jul 2015, 05:46

Five girls can stand in a line in 5! = 120 ways.

Let M(Maggie) and L(Lisa) be treated as a single person ML. Now, ML can be placed in -x-x-x- any one of the 4 empty slots in 4! = 24 ways. ML can be ordered between themselves in 2! ways. So, the number of ways ML can stand together = 24 * 2 = 48.

So, the number of ways they don't stand together is 120 - 48 = 72 ways. Ans (D). _________________

I used to think the brain was the most important organ. Then I thought, look what’s telling me that.

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...