Arrangements Question

10 seats in a row are available for Alice and 9 friends.
(a) All possible arrangements =10! = 3,628,800 No problem

(b)(i) Only 5 friends out of the 9 turn up, how many ways can they all be seated in the 10 available seats.
10P6 = 151,200 No problem

(ii) Find the permutations for any two of her friends (F) sit next to Alice(A).
FAF_ _ _ _ _ _ _ so there are 7 possible position outcomes

I guess part of the solution is using the value found in part (b).
Don't know what to do now!

ans =33,600

(edited 3 months ago)

Reply 1

tonyiptony

For me, often times when it comes to combinations, you want to subdivide your task into steps, count how many possible ways in each step, then multiply everything.

So here, one possible "directions" for the folks to sit is:
(1) Alice pick a seat first - she's the most annoying to deal with;
(2) Choose two friends to sit next to Alice;
(3) Seat the two friends down;
(4) Seat the remaining three folks.
(Verify) No double counting.

See if this line of subdividing the task helps.