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The host of a dinner party must determine how to seat himself and 5 gu

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The host of a dinner party must determine how to seat himself and 5 gu  [#permalink]

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New post 12 Dec 2018, 03:06
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A
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C
D
E

Difficulty:

  5% (low)

Question Stats:

77% (00:47) correct 23% (01:10) wrong based on 82 sessions

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The host of a dinner party must determine how to seat himself and 5 gu  [#permalink]

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New post 12 Dec 2018, 03:11
Bunuel wrote:
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


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  [#permalink]

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New post 02 Feb 2019, 05:58
Bunuel wrote:
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


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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Re: The host of a dinner party must determine how to seat himself and 5 gu  [#permalink]

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New post 03 Feb 2019, 00:58
The Host also has 6options to rotate-120*6 =720?

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Re: The host of a dinner party must determine how to seat himself and 5 gu  [#permalink]

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New post 04 Feb 2019, 10:05
Bunuel Why are we not considering that the host has 6 choices himself and doing 5! X 6?
I am not clear with the concept.
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Re: The host of a dinner party must determine how to seat himself and 5 gu  [#permalink]

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New post 06 Feb 2019, 19:47
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Bunuel wrote:
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



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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Re: The host of a dinner party must determine how to seat himself and 5 gu   [#permalink] 06 Feb 2019, 19:47
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