Kate, Leonard, and 3 other schoolchildren

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Kate, Leonard, and 3 other schoolchildren are to be seated in a row of 5 seats facing their class. For how many of the possible arrangements of the 5 children in the 5 seats will either Kate or Leonard be seated in the middle seat?

A. 5
B. 10
C. 25
D. 32
E. 48

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by Bill@VeritasPrep » Thu Nov 29, 2012 10:51 am
If we put Kate in the middle seat, we have 4 seats left for 4 schoolchildren. 4! = 24.

If we put Leonard in the middle seat, we still have 4 seats for the 4 remaining children, so another 4!=24, which means we have a total of 48 arrangements with either Kate or Leonard in the middle seat.
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by Brent@GMATPrepNow » Thu Nov 29, 2012 1:33 pm
ritumaheshwari02 wrote:Kate, Leonard, and 3 other schoolchildren are to be seated in a row of 5 seats facing their class. For how many of the possible arrangements of the 5 children in the 5 seats will either Kate or Leonard be seated in the middle seat?

A. 5
B. 10
C. 25
D. 32
E. 48
Another approach is to recognize that we can seat the 5 children in 5! ways (120 ways).

With all things considered equal, each person is equally likely to sit in the middle chair.
So, of those 120 arrangements, 1/5 will feature Kate in the middle, and 1/5 will feature Leonard in the middle.
In other words, 2/5 of the 120 arrangements will feature either Kate or Leonard in the middle seat.

2/5 of 120 is 48, so the correct answer is E

Cheers,
Brent
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by Bill@VeritasPrep » Thu Nov 29, 2012 1:56 pm
Brent@GMATPrepNow wrote:
ritumaheshwari02 wrote:Kate, Leonard, and 3 other schoolchildren are to be seated in a row of 5 seats facing their class. For how many of the possible arrangements of the 5 children in the 5 seats will either Kate or Leonard be seated in the middle seat?

A. 5
B. 10
C. 25
D. 32
E. 48
Another approach is to recognize that we can seat the 5 children in 5! ways (120 ways).

With all things considered equal, each person is equally likely to sit in the middle chair.
So, of those 120 arrangements, 1/5 will feature Kate in the middle, and 1/5 will feature Leonard in the middle.
In other words, 2/5 of the 120 arrangements will feature either Kate or Leonard in the middle seat.

2/5 of 120 is 48, so the correct answer is E

Cheers,
Brent
That's one of my favorite things about combinatorics questions. There are usually multiple good ways to solve a given problem, so it's more a matter of thinking your way through a problem then memorizing strict rules.
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by Brent@GMATPrepNow » Thu Nov 29, 2012 3:41 pm
Bill@VeritasPrep wrote:
Brent@GMATPrepNow wrote:
ritumaheshwari02 wrote:Kate, Leonard, and 3 other schoolchildren are to be seated in a row of 5 seats facing their class. For how many of the possible arrangements of the 5 children in the 5 seats will either Kate or Leonard be seated in the middle seat?

A. 5
B. 10
C. 25
D. 32
E. 48
Another approach is to recognize that we can seat the 5 children in 5! ways (120 ways).

With all things considered equal, each person is equally likely to sit in the middle chair.
So, of those 120 arrangements, 1/5 will feature Kate in the middle, and 1/5 will feature Leonard in the middle.
In other words, 2/5 of the 120 arrangements will feature either Kate or Leonard in the middle seat.

2/5 of 120 is 48, so the correct answer is E

Cheers,
Brent
That's one of my favorite things about combinatorics questions. There are usually multiple good ways to solve a given problem, so it's more a matter of thinking your way through a problem then memorizing strict rules.
Agreed! In fact we could probably apply that sentiment to most GMAT math questions. There's almost always more than 1 approach that works.

Cheers,
Brent
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