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.
How many ways 3 persons A, B , C can be put in a row of six empty chair such that A nd B don’t sit together
combinatorial-proofscombinatoricscontest-math
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Best Answer
Your reasoning is correct.
Herer another way:
All together:
$$\color{blue}{\binom 63 \cdot 3! - 2\cdot 5 \cdot 4 = 80} $$