(a) Find a recurrence relation for the number of ways to arrange three types of flags on a flagpole n feet high: red flags (1 foot high), gold flags (1 foot high),and green flags (2 feet high).

(b) Repeat part (a) with the added condition that there may not be three 1-foot flags (red or gold) in a row.

(c) Repeat part (a) with the condition of no red above gold above green (in a row).

I need a detailed process

If the last flag used is a red flag, then the other flags are n-1 foot high, so they can be seen as arranged on a smaller flagpole of n-1 feet high, which can be done in [tex]f_{n-1}[/tex] ways.

Similarly, If the last flag used is a gold flag, then the other flags can be seen as arranged on a smaller flagpole of n-1 feet high. This can be done in [tex]f_{n-1}[/tex] ways.

If the last flag used is green, the other flags are n-2 feet high, so the flagpole can be arranged in [tex]f_{n-2}[/tex] ways.

Using the sum rule, we obtain that [tex]f_n=2f_{n-1}+f_{n-2}[/tex] for all n≥3. Listing all the combinations of flags, the initial conditions are [tex]f_1=2, f_2=5[/tex].

b) If the last flag used is green, there are [tex]f_{n-2}[/tex] ways to choose the other flagd.

If the last flag used is not green, it is a 1 ft flag. It can happen that a green flag was used before, then there are [tex]2f_{n-3}[/tex] arrangements (counting red and gold). If not, then another 1ft flag was used before (with 4 possible combinations). To satisfy the condition, the flag before those two must be green, and the remaining flags can be chosen in [tex]4f_{n-4}[/tex] ways.

By the sum rule, [tex]f_n=f_{n-2}+2f_{n-3}+4f_{n-4}[/tex] for all n≥4.

c) If the last flag used is gold, the condition is always satisfied no matter what flags are used below, then there are [tex]f_{n-1}[/tex] ways to arrange the flagpole.

Similarly, If the last flag used is green, there are [tex]f_{n-2}[/tex] ways to arrange the flagpole.

If the last flag used is red, the arrangement of n-1 flags below can't use (gold green) as the last flags. Call this kind of arrangement "bad". The number of bad arrangements is [tex]f_{n-3}[/tex] (3 ft are fixed gold green) so the number of valid arrangements is [tex]f_{n-1}-f_{n-3}[/tex].

Using the sum rule, [tex]f_{n}=2f_{n-1}+f_{n-2}-f_{n-3}[/tex] for all n≥3.