Combinatorics: Finding recurrence relation

In summary: The initial values are a_1=2, a_2=4, a_3=1, a_4=4, a_5=12, a_6=21, a_7=30, a_8=24, a_9=-12. In summary, the number of ways to arrange three types of flags on an n foot flag pole, with one condition of not having three 1 foot flags in a row, can be represented by the recurrence relation a_n = a_(n-3)+2*a_(n-4)+4*a_(n-5) and the initial values a_1=2, a_2=4, a_3=1, a_4=4
  • #1
Askhwhelp
86
0
Find the number of ways to arrange three types of flags on an n foot flag pole: red flags (1 foot high), white flags (1 foot high), blue flags (3 feet high)

Find a recurrence relation for this number with one condition that there cannot be three 1 foot flags in a row (regardless of their color).

R= Red, W=White, B=Blue

a_1 = 2, R, W
a_2 = 4, RW, WR, WW, RR
a_3 = 1, B
a_4 = 4, B(a_1), W(a_3), R(a_3)
a_5 = 12, B(a_2), W(a_4), R(a_4)
a_6 = 21, B(a_3), W(a_5-2(a_3)), R(a_5-2(a_3))
a_7 = 30, B(a_4), W(a_6-2(a_4)), R(a_6-2(a_4))
a_8 = 24, B(a_5), W(a_7-2(a_5)), R(a_7-2(a_5))
a_9 = -number,B(a_6), W(a_8-2(a_6)), R(a_8-2(a_6))
Since there is a negative number, I do not think this is right...could anyone point out what I did wrong?
 
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  • #2
R(a_5-2(a_3))

doesn't look like the right way to count the number of flags that are six high and have a red on top. For starters, you are claiming there are ten such flags, and really you can only do (I think)

R
R/W
B
R/W

or
R
B
R/W
R/W

of which there are only eight such flags. Instead of trying to subtract away flags that you originally constructed that are actually illegal, just think about if you start with a R on top, what are the only possible chains you can have before you reach a B? Once you know that you have placed B everything underneath it is counted using your recurrence relation.
 
  • #3
I approach this sort of problem by creating several related recurrence terms. You can have the total number of height n, the number of height n with at most one 1-foot flag on top, and the number that have a 3-foot flag on top. Then you can construct three recurrence relations that connect them.
 
  • #4
Let a_n be the arrangement of flags at n feet. Just figure it out...is it a_n = a_(n-3)+2*a_(n-4)+4*a_(n-5). The initial value is a_2= 1, a_2 = 4, a_3 = 1, a_4 = 4,a_5 = 12
 
  • #5
Askhwhelp said:
Let a_n be the arrangement of flags at n feet. Just figure it out...is it a_n = a_(n-3)+2*a_(n-4)+4*a_(n-5). The initial value is a_2= 1, a_2 = 4, a_3 = 1, a_4 = 4,a_5 = 12
I get the same recurrence relation.
 

1. What is combinatorics and why is it important?

Combinatorics is a branch of mathematics that studies the ways in which objects can be arranged or selected. It is important because it has applications in various fields such as computer science, statistics, and physics.

2. What is a recurrence relation in combinatorics?

A recurrence relation is a mathematical equation that describes the relationship between the terms in a sequence. In combinatorics, it is used to find the number of possible combinations or arrangements of a given set of objects.

3. How do you find a recurrence relation?

To find a recurrence relation, you first need to identify the pattern in the sequence of terms. Then, you can express this pattern as a mathematical equation using the previous terms in the sequence. This equation will be the recurrence relation for the given problem.

4. What is the difference between a recursive and explicit recurrence relation?

A recursive recurrence relation is one in which the current term is defined in terms of previous terms in the sequence, while an explicit recurrence relation is one in which the current term is defined directly using a formula. Recursive relations are useful for solving problems with smaller inputs, while explicit relations are useful for larger inputs.

5. How is a recurrence relation used to solve combinatorics problems?

A recurrence relation is used in combinatorics to find the number of combinations or arrangements of a given set of objects. By solving the recurrence relation, we can determine the number of possibilities for a particular scenario, which can then be used to solve more complex combinatorics problems.

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