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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:

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?

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:

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 _________________

_ _ _ _ _ 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

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