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:

Five students Joe, Katy, Lori, Michael and Natasha must [#permalink]

Show Tags

23 Jul 2006, 00:08

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Five students Joe, Katy, Lori, Michael and Natasha must stand in a line. Natasha cannot stand at either the front or back and Lori must stand in the middle. how many more combinations would there be if Lori could stand anywhere in line?

a) 12
b) 30
c) 48
d) 60
e) 72

when i tried to solve this i got:

33121 = 18 (with Lori standing in the middle)

44321 = 96 (with Lori standing anywhere)

96 - 18 = 78 isnt even an answer choice. where did i go wrong?

Five students Joe, Katy, Lori, Michael and Natasha must stand in a line. Natasha cannot stand at either the front or back and Lori must stand in the middle. how many more combinations would there be if Lori could stand anywhere in line?

a) 12 b) 30 c) 48 d) 60 e) 72

when i tried to solve this i got:

33121 = 18 (with Lori standing in the middle)

44321 = 96 (with Lori standing anywhere)

96 - 18 = 78 isnt even an answer choice. where did i go wrong?

Red part is wrong.

Second position can be occupied by 4 (All except Lori)
So it will be 3*4*1*2*1 = 24
_________________

If Lori takes the middle, then Natasha has 2 places, and the others have 3 x 2 x 1 places. In total, 12.

If Lori can go anywhere, then Natasha has 3 places, the others have 4 x 3 x 2 x 1 places. In total, 72.

Increase = 72 - 12 = 60
_________________

Uh uh. I know what you're thinking. "Is the answer A, B, C, D or E?" Well to tell you the truth in all this excitement I kinda lost track myself. But you've gotta ask yourself one question: "Do I feel lucky?" Well, do ya, punk?

_ _ _ _ _ 5 spots; plug L in the middle, so N has 2 possible spots. For each of her spots there are 3p3 possibilities for the rest. So combos are
2 x 3p3=12

next make L a little less fussy. Now N has 3 possible spots. For each position she chooses, the other 4 can have 4p4 arrangements
=3 x 4p4 = 3 x 24 = 72

With Lori in middle,total permutations to 4! (lori is in fixed position)
perms where Lori in middle, Natasha could be in back or front =3! * 2
(lori is in fixed position and Natasha has two options)
total where Lori in middle, Natasha not in back or front = 4! - (3! * 2) = 12

With Lori any where , total permutations = 5!
perms where natasha in back or front = 4! * 2
total where Natasha not in back or front = 5! - (4! * 2) = 120 - 48 = 72