How many ways 3 persons A,B and C can seat in a row of six empty chairs such that A and B don't sit together?
My Attempt
Let us introduce three more persons D, E and F. Then these six persons can be placed in six chair in $6!$ ways. Now in these placements if we tie together A and B then number of permutation where A and B seat together is $2 \cdot 5!$ ways. Hence there are $(6! -2\cdot 5!)$ permutations with six persons which do not have A nd B together. Now we remove D , E , F to create six chairs and three persons A , B and C seating with condition A and B are not seating together. But empty chairs are indistinguishable and D , E , F can be permuted in $3!$ ways. Hence final answer is $\frac{(6! -2\cdot 5!)}{3!}=80$.
Is my answer and logic correct? Please verify. Thanks in advance.

  • $\begingroup$ Does $A$ and $B$ seating together mean they are adjacent? For example, does $- A -BC-$ count as $A$ and $B$ seating together? $\endgroup$ – Thomas Bladt Nov 6 '18 at 5:10
  • $\begingroup$ $-A-BC--$ is acceptable but $ABC---$ or $C-BA--$ are not acceptable. $\endgroup$ – rugi Nov 6 '18 at 5:12

Your reasoning is correct.

Herer another way:

  • When you bind $A,B$ together you can seat them in $5$ ways beside each other on $6$ chairs and permute the two: $\color{blue}{2\cdot 5}$
  • From the remaining $4$ chairs you choose one for $C$: $\color{blue}{4}$
  • All possible ways of placing $3$ onto the $6$ chairs (choose $3$ seats from $6$ and arrange $A,B,C$ in them): $\color{blue}{\binom 63 \cdot 3!}$

All together:

$$\color{blue}{\binom 63 \cdot 3! - 2\cdot 5 \cdot 4 = 80} $$

  • $\begingroup$ Thanks for pointing calculation mistake and another nice alternative. $\endgroup$ – rugi Nov 6 '18 at 5:17

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