Question

Mrs. Teabottle is quite worried about the annual Gardenia Banquet. Ms. Aggravant, Mr. Bellicose, Mrs. Colic and Dr. Dreck all detest each other and will have to sit at different tables. Fortunately, there are five tables (Red, Green, Blue, Yellow and Orange) to choose from. But neither Ms. Aggravant and Mr. Bellicose may sit at the Red or Green Tables. Mrs. Colic absolutely cannot be allowed to go near the Yellow or Orange tables (we all remember what happened last year!) and Dr. Dreck must not sit at the Blue or Yellow tables. Oh dear! cries Mrs. Teabottle. But she worries too much! How many ways can she seat these difficult guests

Answer 1

Answer:

20

Step-by-step explanation:

To abbreviate, I will refer to each person by their initials (A,B,C,D).

We will count all the possibilities depending on C and completing each case.

C can sit on the blue, red or green tables. Suppose that C sits on the blue table. Now, A can sit on the yellow or orange tables, then there are 2 possibilities to sit A. After choosing, there remains only one choice for B, either the yellow or orange table, the one which A didn't sit.

Now, the blue, yellow and orange tables are occupied, so D can sit on the red or green tables. Hence, we have 2 choices for D. In total, we have 2×2=4 ways to sit everyone if we choose blue for C.

Suppose that C sits on the green or red tables (2 choices). Now, we split this in two cases.

If we use the yellow and blue tables for A and B, there are 2 ways of seating them (A to yellow, B to blue, or A to blue, B to yellow). For D, we have 2 choices, the red/green table (depending on C) and the orange table. Thus, in this case, we have 2×2×2=8 ways of seat the guests.

If we don't use precisely the yellow and blue tables for A and B, one of them must sit at the orange table. There are 2 ways of doing this (A in orange or B in orange). The other must sit on the blue or yellow tables (2 choices). Finally, D has only 1 choice, the table not used by C. Then we have 2×2×2=8 ways of doing this.

Therefore, considering all the cases we have 4+8+8=20 ways of seating the guests.