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jim441
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There are five chairs in a room {C1 C2 C3 C4 C5} and there a Updated on: 26 Apr 2023, 19:49
Originally posted by jim441 on 25 Apr 2023, 11:16.
Last edited by jim441 on 26 Apr 2023, 19:49, edited 1 time in total.
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Q1 There are five chairs in a room {C1 C2 C3 C4 C5} and there are five people ( A,B,C,D,E). In how many ways these five people can be seated on these chairs such that D & E cannot sit on C4 and C5.

Q2 same question but this time the chairs are kept around a circular table.
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There are five chairs in a room {C1 C2 C3 C4 C5} and there a 26 Apr 2023, 16:56
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Hi Jim.

Moving this question into the problem-solving sub forum.

PS. Do you happen to have the answer choices and the source of this question?

Posted from my mobile device
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jim441
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There are five chairs in a room {C1 C2 C3 C4 C5} and there a 26 Apr 2023, 19:49
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Hi Jim.

Moving this question into the problem-solving sub forum.

PS. Do you happen to have the answer choices and the source of this question?

Posted from my mobile device
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Hi bb, This is a self made question. I was reading combinations/probability and got this doubt so asked it here.
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There are five chairs in a room {C1 C2 C3 C4 C5} and there a 27 Apr 2023, 07:39
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jim441
Q1 There are five chairs in a room {C1 C2 C3 C4 C5} and there are five people ( A,B,C,D,E). In how many ways these five people can be seated on these chairs such that D & E cannot sit on C4 and C5.

Q2 same question but this time the chairs are kept around a circular table.
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So I would break this into the two blocks of seats.

FIRST let's think about how many options we have for C1, C2, and C3.

D an E must be two of the three selected for these, along with either A, B, or C.

The possible arrangements of these 3 people will be 3! = 6.

BUT since there are three choices for the whoever that third person is, we'll multiply that by 3.

So there are 18 possible arrangements of C1, C2, and C3.

After these choices are made, there are only 2 choices for the second block of C4 and C5.

So in all, there are 2*18 = 36 possible arrangements.

Alternatively, you could think about your choices for seats C4 and C5. You have three choices for C4 and then two choices for C5, resulting in 6 choices for those chairs. Then, for the first block, you'd have three choices for C1, two for C2, and one for C3, also resulting in 6 choices.

So you have 6*6 = 36 total options.

I don't see how putting these chairs in a circle changes the situation. I think you're slightly blending question types here, trying to bring in situations where the seating in a circle causes 'duplications' that a straight line does not. But the restrictions of D and E on C4 and C5 actually turn the 'circle' back into a straight line problem.
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