sorry again, really bad at this topic. (1 Viewer)

[yashbb]

four boys and four girls are arranged in a circle. in how many ways can they be arranged if two particular boys do not wish to sit next to each other

 

[B1andB2]

For this one, you would follow the same steps as the previous question BUT minus it from the unrestricted total

Complementary event: total which is (8-1)!- ways(they sit together)

so (8-1)! - (2! x (7-1)!)

 

[icycledough]

Not sure if this is completely the right intuition (been a while since I did perms and comms):

The first boy has a fixed position (so has 1 option; wherever you choose).

The other boy has 2 places he cannot sit (either side of the first boy); so has 8 (total) - 2 (either side of 1st boy) - 1 (first boy's spot) = 5 spots.

Remaining 6 people have 6! ways they can arrange themselves.

So 1 x 5 x 6! = 3600 ways.

Could you let me know if this is the right answer ... I'm not 100% with my reasoning.

 

[yashbb]

For this one, you would follow the same steps as the previous question BUT minus it from the unrestricted total

Complementary event: total which is (8-1)!- ways(they sit together)

so (8-1)! - (2! x (7-1)!)

Thank you so much, i cannot covey how grateful i am for this.

 

[yashbb]

Not sure if this is completely the right intuition (been a while since I did perms and comms):

The first boy has a fixed position (so has 1 option; wherever you choose).

The other boy has 2 places he cannot sit (either side of the first boy); so has 8 (total) - 2 (either side of 1st boy) - 1 (first boy's spot) = 5 spots.

Remaining 6 people have 6! ways they can arrange themselves.

So 1 x 5 x 6! = 3600 ways.

Could you let me know if this is the right answer ... I'm not 100% with my reasoning.

Thanks so much, really appreciate all of this help

 

[B1andB2]

Not sure if this is completely the right intuition (been a while since I did perms and comms):

The first boy has a fixed position (so has 1 option; wherever you choose).

The other boy has 2 places he cannot sit (either side of the first boy); so has 8 (total) - 2 (either side of 1st boy) - 1 (first boy's spot) = 5 spots.

Remaining 6 people have 6! ways they can arrange themselves.

So 1 x 5 x 6! = 3600 ways.

Could you let me know if this is the right answer ... I'm not 100% with my reasoning.

yep, that's the right answer, just a different way of doing it

 

[icycledough]

Thanks so much, really appreciate all of this help

Don't worry ... I checked B1andB2 answer and it ends up also being 3,600 but with a different style of approaching it. Either way, they both get the answer, so that's all that matters

 

[yashbb]

Not sure if this is completely the right intuition (been a while since I did perms and comms):

The first boy has a fixed position (so has 1 option; wherever you choose).

The other boy has 2 places he cannot sit (either side of the first boy); so has 8 (total) - 2 (either side of 1st boy) - 1 (first boy's spot) = 5 spots.

Remaining 6 people have 6! ways they can arrange themselves.

So 1 x 5 x 6! = 3600 ways.

Could you let me know if this is the right answer ... I'm not 100% with my reasoning.

also sir you have got the correct answer.

 

[CM_Tutor]

That the answers are the same can be confirmed by noting that

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