Sorry. Chasing not allowed

Probability Level 4

In Disneyland, Florida on the occasion of New Year a group of 10 children took a ride of a small merry-go-round.

Initially $10$ children (say, $A,B,C,D,E,F,G,H,I,J$ ) sitting one behind the other in a $10$ seater merry-go-round. Later, they decide to switch seats so that each child has a new companion in front in other words, $J$ shall not have $A$ in front, $A$ shall not have $B$ in front, $C$ shall not have $D$ in front and so forth.

In how many ways can this arrangement be done?

Conclusion: The seats are identical.

The answer is 120288.

1 solution

Alapan Das
Jan 10, 2020

This is problem goes similar to the disarrangement problem. Consider that seats are not identical. Seat number matters.

Total number of possible arrangement is $10!$ .
Now, choosing one child, say $A$ and saying it is chaser means in front of $A$ , $B$ is seated. So, the number of arrangements with such $1-chaser$ is $\binom{10}{1} ×10×8!$
Number of arrangements with such $2-chaser$ is $\binom{10}{2} ×10×7!$ ..

So, our required expression is

$10!-\binom{10}{1} ×10×8!+\binom{10}{2} ×10×7!-......\binom{10}{8} ×10×1-\binom{10}{9}×10+\binom{10}{10} ×10=1202880$ .

We assumed that the seats are not identical for symmetry, but actually they are given to be identical.

So, we have to divide this obtained number by $10$ as all $10$ seats are identical. Hence, the number of such no-chasing arrangements is $120288$ .

Sorry. Chasing not allowed

The answer is 120288.

1 solution

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