Intereting Posts

ACC on principal ideals implies factorization into irreducibles. Does $R$ have to be a domain?
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Symmetry of Grassmanians
Intersection maximal ideals of a polynomial ring
Is $f^{-1}(f(A))=A$ always true?
Are all measure zero sets measurable?
Halmos, Naive Set Theory, recursion theorem proof: why must he do it that way?
Changing a bezier curve by dragging a point on the curve itself rather than a control point
Drawing large rectangle under curve
Reference for a “wild” problem
Why is $dy dx = r dr d \theta$
the sum of powers of $2$ between $2^0$ and $2^n$
Product of sums of square is a sum of squares.
central limit theorem for a product
Confusion regarding what kind of isomorphism is intended.

This picture was in my friend’s math book:

Below the picture it says:

- How many 5-cards poker hands are there containing at least 3 of the 4 suits?
- Probability of drawing exactly 13 black & 13 red cards from deck of 52
- Number of ways to partition a rectangle into n sub-rectangles
- $p$-Splittable Integers
- 2 regular graphs and permutations
- Expected value of number of edges of a connected graph

There are $3072$ ways to draw this flower, starting from the center of

the petals, without lifting the pen.

I know it’s based on combinatorics, but I don’t know how to show that there are actually $3072$ ways to do this. I’d be glad if someone showed how to show that there are exactly $3072$ ways to draw this flower, starting from the center of the petals, without lifting the pen (assuming that $3072$ is the correct amount).

- Prove by Combinatorial Argument that $\binom{n}{k}= \frac{n}{k} \binom{n-1}{k-1}$
- Distinct digits in a combination of 6 digits
- Prove $\sum_{j=0}^n \left(-\frac{1}{2}\right)^j \binom{n}{j}\binom{n+j}{j}\binom{j}{k} = 0$ when $n+k$ is odd
- Software for drawing and analyzing a graph?
- Probability for the length of the longest run in $n$ Bernoulli trials
- A question on calculating probabilities for the random walk
- How do I tackle this combinatorics problem about married couples around a table?
- Expected size of subset forming convex polygon.
- Lower bound for monochromatic triangles in $K_n$
- Min. number of vertices in graph as function of $\kappa(G)$ and $\operatorname{diam}(G)$

First you have to draw the petals. There are $4!=24$ ways to choose the order of the petals and $2^4=16$ ways to choose the direction you go around each petal. Then you go down the stem to the leaves. There are $2! \cdot 2^2=8$ ways to draw the leaves. Finally you draw the lower stem. $24 \cdot 16 \cdot 8=3072$

At the beginning you could go 8 different ways, then you could go 6 different ways, then you could go 4 and 2 different ways but in the down of the picture you could go at first 4 different ways and 2 at the end.

$8\cdot6\cdot4\cdot2\cdot4\cdot2 = 3072$

Looking at the picture, there are 4 phases.

- Draw the petals
- Draw the upper stem
- Draw the leaves
- Draw the lower stem

Lets label these $A,B,C,D$. Clearly, the total number of ways to draw the flower is simply;

$$Total = A \times B \times C \times D$$

We can see that the upper and lower stem are un-ambiguous; i.e. there is only one way to draw them. Thus $B=D=1$. So our equation becomes

$$Total = A \times C$$

Now lets look at the leaves first. The factors in it are:

a. Which leaf do you draw first?

b. What direction do you use for the first leaf?

c. What direction do you use for the second leaf?

There are 2 possibilities for each, so $C=2\times 2 \times 2=8$.

Now lets look at the petals.

a. Which petal do you draw first? (4 choices)

b. Which petal do you draw second? (3 choices)

c. Which petal do you draw third? (2 choices)

d. Which direction do you draw the X petal? (2 choices for each petal)

So $A=4 \times 3 \times 2 \times 1 \times 2^4=24 \times 16=384$.

And

$$Total=A \times C = 384 \times 8 = 3072$$

Think through each of the “decision points,” and think through how many options you have at each decision point.

Starting in the middle of the flower, there are 4 petals you could choose to draw first. For each petal, there are 2 ways to draw it: forwards or backwards. So you have 8 options. Draw your first petal, and now there are 6 options of what to do next (3 petals, 2 ways to draw each). And so on. So you have 8*6*… Do you see what to do next?

Others have already solved the problem. I just wanted to add that, properly speaking, this problem and/or its solution belong to a branch of math called graph theory. Formally:

Proposition.The following undirected graph has $3072$ directed Eulerian trails starting at $x$ and ending at $z$.

We could be even more precise by describing the graph in more formal terms, of course, rather than just drawing a picture. This can be done by giving a symmetric function $\{x,y,z\}^2 \rightarrow \mathbb{N}$ that tells us how many edges go between any two vertexes. In this particular case, our function is

Enumerate the possibilities:

- Draw one of 4 top petals, in one of two directions: 4*2
- Draw one of 3 remaining top petals, in one of two directions: 3*2
- Draw one of 2 remaining top petals, in one of two directions: 2*2
- Draw last petal in one of two directions: 2
- Draw one of 2 leaves, in one of two directions: 2*2
- Draw remaining leaf, in one of two directions: 2

Thus, we have the total number of possibilities:

(4*2)(3*2)(2*2)(2)(2*2)(2)=3072

The number of possible ways to draw 4 petals in order is $4!$. Then let’s imagine that if the drawing of the petal is clockwise then the petal is black and if the drawing is counterclockwise the petal is white. So the drawing can be: $0000, 0001, … 1111$ therefore $2^4$ different drawings $0$ for black $1$ for white. Therefore $4! \cdot 2^4$ ways to draw the upper petals. Similary $2! \cdot 2^2$ ways to draw the lower petals and yes, $4! \cdot 2^4 \cdot 2! \cdot 2^2 = 3072$ It is pretty much @Ross Millikan’s solution I just added the coloring ilustration.

EDIT: I have only just noticed that @openspace has already supplied an answer using this idea, so all credit must go to him for being the first one to think of this method. I shall leave my answer though as it seems to flesh out the idea.

An alternative approach:

As others have noted you obviously have to draw all the petals first, then go down the stem and draw all the leaves before drawing the end of the stem. Starting in the centre of the petal there’s a possible of $8$ lines in which to start drawing the petals. After you pick one then there’s $6$ possible ways to pick the line for drawing the second petal, then $4$, then finally $2$ to pick how to draw the last petal.

You then move on down the stem (only $1$ way) to the centre of the leaves. There’s $4$ lines to pick from to start drawing the leaves, and then after drawing the first leaf there’s only $2$ lines to pick from to draw the last leaf. Then there’s drawing the end of the stem (only $1$ way). So in total there are $8\cdot 6\cdot 4\cdot 2 \cdot 1\cdot 4\cdot 2\cdot 1=3072$ ways to draw the flower as desired.

**The presumption is wrong!**

Most answers here start like this:

“First you have to draw the petals.”

This is a mistake. You could start drawing the flower from two points: The centerpoint of the blossom OR the endpoint at the lower stem.

Result is:

There are **2 x 3072 = 6144** ways to draw the Flower!

I think it might have to do with the way you asked the question. ðŸ˜‰

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