Mathematics, 24.12.2019 17:31 BeeShyanne

(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

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Answer from: nsgtritent

(See detailed process below)

Step-by-step explanation:

Let f_n be the number of ways of arranging such flagpole with the given conditions.

a) When arranging a flagpole of n feet high, consider the following cases

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 $f_{n-1}$ 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 $f_{n-1}$ ways.

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

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

b) If the last flag used is green, there are $f_{n-2}$ 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 $2f_{n-3}$ 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 $4f_{n-4}$ ways.

By the sum rule, $f_n=f_{n-2}+2f_{n-3}+4f_{n-4}$ 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 $f_{n-1}$ ways to arrange the flagpole.

Similarly, If the last flag used is green, there are $f_{n-2}$ 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 $f_{n-3}$ (3 ft are fixed gold green) so the number of valid arrangements is $f_{n-1}-f_{n-3}$ .

Using the sum rule, $f_{n}=2f_{n-1}+f_{n-2}-f_{n-3}$ for all n≥3.

Answer from: Quest

false

step-by-step explanation:

4y - 6 + 4y^2 - 9

before simplifying this contains 4 terms

after simplifying

4y -15 + 4y^2

it contains 3 terms

Answer from: Quest

mental cunning

explanation:

he is often referred to as a man of twists and turns

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(a) find a recurrence relation for the number of ways to arrange three types of flags on a flagpole...