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Bunuel
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The host of a dinner party must determine how to seat himself and 5 gu 12 Dec 2018, 03:06
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D
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5% (low)

Question Stats:

79% (00:45) correct 21% (01:09) wrong based on 196 sessions

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The host of a dinner party must determine how to seat himself and 5 guests in a single row. How many different seating arrangements are possible if the host always chooses the same seat for himself?

A. 6
B. 15
C. 21
D. 120
E. 720
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D
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The host of a dinner party must determine how to seat himself and 5 gu 12 Dec 2018, 03:11
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Bunuel
The host of a dinner party must determine how to seat himself and 5 guests in a single row. How many different seating arrangements are possible if the host always chooses the same seat for himself?

A. 6
B. 15
C. 21
D. 120
E. 720
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Host sits on same seat so other guests have an option of
total ways of sitting 5! * 1= 120 IMO D
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The host of a dinner party must determine how to seat himself and 5 gu 02 Feb 2019, 05:58
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Bunuel
The host of a dinner party must determine how to seat himself and 5 guests in a single row. How many different seating arrangements are possible if the host always chooses the same seat for himself?

A. 6
B. 15
C. 21
D. 120
E. 720
Show more

6 seats total but the host always sits in the same seat so we are only concerned with the other 5.

If we go through each seat and assume the host sits in the first one (although his position doesn't matter) we can work out the answer.

There are 5 guests that could sit in the first seat, after that there are 4 remaining guests that could sit in the second seat, 3 in the next, etc.

5! = 120
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The host of a dinner party must determine how to seat himself and 5 gu 03 Feb 2019, 00:58
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The Host also has 6options to rotate-120*6 =720?

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The host of a dinner party must determine how to seat himself and 5 gu 06 Feb 2019, 19:47
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Bunuel
The host of a dinner party must determine how to seat himself and 5 guests in a single row. How many different seating arrangements are possible if the host always chooses the same seat for himself?

A. 6
B. 15
C. 21
D. 120
E. 720
Show more


Since he always chooses the same seat, the remaining 5 guests can be seated in 5! = 120 ways.

Note: The wording of the question is confusing, and you might think that the answer should be 6! = 720.

The fine distinction between the incorrect solution and the correct one is that, while there are 6 individuals to be seated, we don’t consider the host as one of them. Here are some examples that illustrate the phrase “the host always chooses the same seat for himself.”

Example 1: The host chooses the first seat for himself. Thus, there are 5! = 120 ways to seat his guests.

Example 2: The host chooses the second seat for himself. Thus, there are 5! = 120 ways to seat his guests.

We could illustrate this with 4 more examples. In them, the host chooses the 3rd, 4th, 5th, or 6th seat for himself. And in each case, his guests can be seated in 5! = 120 different ways.

We see that, because of the wording of the question, we don’t consider the host’s choice of a seat. Rather, we concentrate on the number of ways he can seat his guests, once he has chosen his seat.

Answer: D
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The host of a dinner party must determine how to seat himself and 5 gu 18 Jul 2023, 12:15
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Hello everyone!
Can anyone be so kind to clarify, -- we don't use the formula for circular permutations (P= (n-1)!), as we have one fixed seat, am I right?

Thansk in a advance!
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The host of a dinner party must determine how to seat himself and 5 gu 18 Jul 2023, 13:10
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stepanyan
Hello everyone!
Can anyone be so kind to clarify, -- we don't use the formula for circular permutations (P= (n-1)!), as we have one fixed seat, am I right?

Thansk in a advance!
Show more

Given that the host's position is fixed, it provides a reference point for the remaining five people. Therefore, these five people can be arranged in 5! ways, not 4!.

This post might also help to clear confusion: https://gmatclub.com/forum/in-how-many- ... l#p3219470
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