25-10-09 Perm Comb seating

Got this idea after one of sinchan's posts...

Not worth a QOD.. but still a nice variation of Seating arrangement.

What is the number of ways to seat people at each of the vertices in both the images below.

Aside: I will try to bring the QOD section back on after december

17 Answers

62
Lokesh Verma ·

no...

1
Bicchuram Aveek ·

Yes there's a symmetry occurring for both the cases which leads to this answer.
Q1 . 2 symmetries
Q2. 4 symmetries.

49
Subhomoy Bakshi ·

similar is the case for 1)11!/2

49
Subhomoy Bakshi ·

2) we can seat the people in 24! ways

but when we see carefully, we see that when we see the arrangement from botton top right and left it gives the same arrangement
so ans is=24!/4

62
Lokesh Verma ·

yup subhomoy and avinav both are correct [1]

1
Avinav Prakash ·

even i am gettin ; 1]11!/2
2]24!/4

1
JOHNCENA IS BACK ·

for q.1>> no. of ways=(5!)2[ considering both hexagons as identical]

49
Subhomoy Bakshi ·

no. 2) 24!/4
considering all angles are vertices....

49
Subhomoy Bakshi ·

11!/2???????

1
xYz ·

is answer for first
2nCn (n-1)! 2

1
Unicorn--- Extinct!! ·

5!4!2! ??

62
Lokesh Verma ·

this one is easy..

no one has done it yet! :(

62
Lokesh Verma ·

no karna..

the answer is slightly simpler.

1
xYz ·

62
Lokesh Verma ·

no not even then :P

* btw now i understand your logic.. and you have made the same mistake that i was thinking that ppl would make [1]

1
xYz ·

sir, if the two hexagons are separated then is my answer right

62
Lokesh Verma ·

there are 2n-1 seats...

n-1 on each ring and one in between..

I did not get your logic... but I dont think that answer is correct

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