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# Five students Joe, Katy, Lori, Michael and Natasha must

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Intern
Joined: 23 Jun 2005
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Five students Joe, Katy, Lori, Michael and Natasha must [#permalink]  23 Jul 2006, 00:08
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:

3 3 1 2 1 = 18 (with Lori standing in the middle)

4 4 3 2 1 = 96 (with Lori standing anywhere)

96 - 18 = 78 isnt even an answer choice. where did i go wrong?
CEO
Joined: 20 Nov 2005
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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Re: standing in line [#permalink]  23 Jul 2006, 00:29
minhthel wrote:
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:

3 3 1 2 1 = 18 (with Lori standing in the middle)

4 4 3 2 1 = 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
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

GMAT Instructor
Joined: 04 Jul 2006
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Quote:
Second position can be occupied by 4 (All except Lori)
So it will be 3*4*1*2*1 = 24

There's a problem with this, if Natasha doesn't go in place two, she must go in place 4
CEO
Joined: 20 Nov 2005
Posts: 2910
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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Kudos [?]: 132 [0], given: 0

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

There's a problem with this, if Natasha doesn't go in place two, she must go in place 4

Got it. Thanks Kevin. I think its time to sleep.

When Lori is NOT in middle:
NXXXX - 4!
XXXXN - 4!
Total = 5! - 2*4! = 72

When Lori is in middle
NXLXX - 3!
XXLXN - 3!
Total = 4! - 2*3! = 12

Final answer = 72-12 = 60
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Director
Joined: 06 May 2006
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D.

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
Director
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Straight D

_ _ _ _ _ 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

difference = 72 -12 = 60
Intern
Joined: 01 Jul 2006
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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

answer = 72 - 12 = 60
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