Probability--suzan

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prachich1987: Legendary Member; Posts: 752; Joined: Sun Sep 12, 2010 2:47 am; Thanked: 20 times; Followed by:10 members; GMAT Score:700

Probability--suzan

by prachich1987 » Tue Feb 01, 2011 5:32 am

In how many ways can you sit 7 people on a bench if Suzan won't sit on the middle seat or on either end?

a) 720
b) 1,720
c) 2,880
d) 5,040
e) 10,080

source : some online material

Thanks!
Prachi

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Source: Beat The GMAT — Problem Solving | Tue Feb 01, 2011 5:32 am

maihuna: Legendary Member; Posts: 1578; Joined: Sun Dec 28, 2008 1:49 am; Thanked: 82 times; Followed by:9 members; GMAT Score:720

by maihuna » Tue Feb 01, 2011 5:39 am

Assuming there is no restriction : :!7 ways

SInce 3 positions are not allowed, allowed position for him is 4, for each of these combo other can be seated in !6 ways

4*!6 = 720*4 = 2880

Charged up again to beat the beast

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pm; Thanked: 127 times; Followed by:10 members

by Night reader » Tue Feb 01, 2011 9:40 am

prachich1987 wrote:In how many ways can you sit 7 people on a bench if Suzan won't sit on the middle seat or on either end?

a) 720
b) 1,720
c) 2,880
d) 5,040
e) 10,080

source : some online material

fairly easy one

let's get to our constraints

--- --- --- --- --- --- ---, in total 7 places and the places # 1,4 and 7 are forbidden for Suzan --> so let the people be seated in these places first, then do whatever they like

No 1 only 6 people (one is out, she's Suzan)
No 4 only 5 people (the same here)
No 7 only 4 people (the same ...)

So we've got --> 6*4*3*5*2*1*4 OR -6-__ -4-__ -3-__ -5-__ -2-__ -1-__ -4-

2,880

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prachich1987: Legendary Member; Posts: 752; Joined: Sun Sep 12, 2010 2:47 am; Thanked: 20 times; Followed by:10 members; GMAT Score:700

by prachich1987 » Tue Feb 01, 2011 10:06 am

Thanks guys
The OA is indeed C

Thanks!
Prachi

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721tjm: Newbie | Next Rank: 10 Posts; Posts: 7; Joined: Thu Jan 20, 2011 4:37 pm; Thanked: 2 times; Followed by:1 members

by 721tjm » Wed Feb 02, 2011 4:30 pm

Usually I'm ok with these but this one threw me for a loop. Could anyone explain the mistake in my approach?

Take the unrestricted combinations and subtract out the restricted combinations

Unrestricted = 7! = 5040 possible combinations

Restriction 1: She won't sit in the first seat.
How many ways can she sit in the first seat? 1*6*5*4*3*2*1 = 6! = 620
Same for middle seat, end seat. So out of 5040 possible combinations, 3*620= 1860 are restricted.

5040 - 1860 = 3180

Thanks a lot

"Any more brain busters??" - Billy Madison

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pm; Location: Milpitas, CA; Thanked: 447 times; Followed by:88 members

by Rahul@gurome » Wed Feb 02, 2011 6:52 pm

721tjm wrote:Usually I'm ok with these but this one threw me for a loop. Could anyone explain the mistake in my approach?

Take the unrestricted combinations and subtract out the restricted combinations

Unrestricted = 7! = 5040 possible combinations

Restriction 1: She won't sit in the first seat.
How many ways can she sit in the first seat? 1*6*5*4*3*2*1 = 6! = 620
Same for middle seat, end seat. So out of 5040 possible combinations, 3*620= 1860 are restricted.

5040 - 1860 = 3180

Thanks a lot

You have wrongly taken 6! as 620.
6! = 720 and not 620.
720*3 = 2160.
So valid combinations are 5040 - 2160 = 2880

Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pm; Thanked: 127 times; Followed by:10 members

by Night reader » Wed Feb 02, 2011 7:02 pm

721tjm wrote:Usually I'm ok with these but this one threw me for a loop. Could anyone explain the mistake in my approach?

Take the unrestricted combinations and subtract out the restricted combinations

Unrestricted = 7! = 5040 possible combinations

Restriction 1: She won't sit in the first seat.
How many ways can she sit in the first seat? 1*6*5*4*3*2*1 = 6! = 620
Same for middle seat, end seat. So out of 5040 possible combinations, 3*620= 1860 are restricted.

5040 - 1860 = 3180

Thanks a lot

you are on right track