How many ways $8$ persons be seated around a circular table facing the center such that $3$ particular persons are always together?
My approach: Bcoz 3 particular persons have to be together, I bundled them into one box. Now before putting then into box I'll first have to select those 3 out of 8 people so it would be done in $$8C3$$ ways. (..am I wrong???) Now there are in all $$(8-3) + 1 = 5 + 1 = 6$$ entities. So these can be arranged across a circular table in $$(6-1)! =5!$$ Ways. And the $3$ people in box can be arranged in $$3!=6$$ ways So total case would be = $$8C3.(5!)(6)$$
Now the problem is: My professor is saying that there won't be any term like $8C3$. And as per my professor the answer is just $$(5!).6$$ But I think we would also have to select first $3$ people out of $8$ right...?(is my thinking right..??) And then perform circular permutation. Like ..wouldn't the selection of $3$ people at the very first place in this problem be a problem that we need to take into consideration...???
Okay if I consider that bcoz problem statement has the term "particular" in it we won't consider 8C3 But plz help some one convincing me if in case I have a term say "any 3 person together"..then would I not have 8C3 in this case..???
Yes answer will be $(5!).3!$ since according to the question they are particular 3 people who will sit together we are not required to select ( selecting any way is meaningless as always any 3 are together ). $$ $$ Considering $3$ to be one unit we need to arrange $(8-3+1)=6$ in $(5!)$ number of ways now the $3$ together can arrange themself in $3!$ number of ways in every arrangement of 6. $$ Total \,arrangements = (5!)(3!)$$