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Martha invited 4 friends to go with her to the movies. There

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Martha invited 4 friends to go with her to the movies. There [#permalink] New post 30 May 2019, 11:21
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Martha invited 4 friends to go with her to the movies. There are 120 different ways in which they can sit together in a row of 5 seats, one person per seat. In how many of those ways is Martha sitting in the middle seat?


[Reveal] Spoiler: OA
24


Math Review
Question: 7
Page: 297
Difficulty: medium

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Re: Martha invited 4 friends to go with her to the movies. There [#permalink] New post 31 May 2019, 04:33
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Carcass wrote:
Martha invited 4 friends to go with her to the movies. There are 120 different ways in which they can sit together in a row of 5 seats, one person per seat. In how many of those ways is Martha sitting in the middle seat?


[Reveal] Spoiler: OA
24


Math Review
Question: 7
Page: 297
Difficulty: medium


Take the task of seating the 5 people and break it into stages.

We’ll begin with the most restrictive stage.

Stage 1: Select someone to sit in the MIDDLE seat
Since we're told Martha MUST sit in the middle seat, we must place her in that seat.
So, we can complete stage 1 in 1 way

Stage 2: Select someone to sit in seat #1
There are 4 people remaining to be seated, so we can complete this stage in 4 ways

Stage 3: Select someone to sit in seat #2
There are 3 people remaining to be seated, so we can complete this stage in 3 ways

Stage 4: Select someone to sit in seat #4
There are 2 people remaining to be seated, so we can complete this stage in 2 ways

Stage 5: Select someone to sit in seat #5
There is 1 person remaining to be seated, so we can complete this stage in 1 way

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus seat all 5 people) in (1)(4)(3)(2)(1) ways (= 24 ways)

Answer: 24

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn

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Re: Martha invited 4 friends to go with her to the movies. There [#permalink] New post 04 Oct 2019, 11:16
Carcass wrote:
Martha invited 4 friends to go with her to the movies. There are 120 different ways in which they can sit together in a row of 5 seats, one person per seat. In how many of those ways is Martha sitting in the middle seat?


[Reveal] Spoiler: OA
24


Math Review
Question: 7
Page: 297
Difficulty: medium


How can we solve this question by permutation? \(\frac{n!}{(n-k)!}\)
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Re: Martha invited 4 friends to go with her to the movies. There [#permalink] New post 04 Oct 2019, 16:14
Expert's post
I do not think you can use that formula.

This is basically a permutation wher the order does not matter.

You do have Martha in the middle but you can choose whatever you wanna of the other four people to sit in the other spots

I hope Brent could shed some more light on this.
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Re: Martha invited 4 friends to go with her to the movies. There [#permalink] New post 04 Oct 2019, 17:13
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Asmakan wrote:
How can we solve this question by permutation? \(\frac{n!}{(n-k)!}\)


There are very few "true" permutation questions on the GRE.

The Fundamental Counting Principal is a far more useful counting strategy.

Cheers,
Brent
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Re: Martha invited 4 friends to go with her to the movies. There [#permalink] New post 13 Oct 2019, 09:13
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Asmakan wrote:
Carcass wrote:
Martha invited 4 friends to go with her to the movies. There are 120 different ways in which they can sit together in a row of 5 seats, one person per seat. In how many of those ways is Martha sitting in the middle seat?


[Reveal] Spoiler: OA
24


Math Review
Question: 7
Page: 297
Difficulty: medium


How can we solve this question by permutation? \(\frac{n!}{(n-k)!}\)



You can use (n-1)! for the round table question , here n = no. of seats i.e. 5

So total number of ways = (5-1)! = 4!
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Re: Martha invited 4 friends to go with her to the movies. There   [#permalink] 13 Oct 2019, 09:13
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