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This picture was in my friend's math book:

Below the picture it says:

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).

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    $\begingroup$ The title appears to be slightly misleading; The starting point is fixed in the plain text. $\endgroup$ – Aki Suihkonen Apr 11 '16 at 11:34
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    $\begingroup$ Meh, without the image the title simply describes the problem. Can't really claim this is "misleading" because there's no emotional expectation to the arbitrary seeming number 3072. Although, I am perplexed the fixed the starting point. Why don't just make the statement there are 6144 ways to draw the flower. $\endgroup$ – fleablood Apr 11 '16 at 17:52
  • $\begingroup$ @fleablood: well, it's reasonable to strip away the obvious symmetry that exists in reversing the order of each path. It's not necessary, but I don't find it perplexing either :-) $\endgroup$ – Steve Jessop Apr 11 '16 at 22:15
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    $\begingroup$ But the symmetry is the entire point of these puzzles. There's only one way to draw it otherwise start at the center, draw each petal, go to th middle, draw each leaf, and finish the stem. $\endgroup$ – fleablood Apr 11 '16 at 22:29
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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. $24 \cdot 16 \cdot 8=3072$

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    $\begingroup$ There's a built in assumption (so far as I can see, unstated) that you may cross a line at a point, but you may not draw back over an existing line (e.g. draw the stem down first and then go back over it to get back to the leaves and petals). $\endgroup$ – Jason Apr 10 '16 at 20:17
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    $\begingroup$ @Jason: yes, that is how I took it. $\endgroup$ – Ross Millikan Apr 10 '16 at 20:19
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    $\begingroup$ @Jason Without this assumption there are infinite ways. If you can draw back and forth, you can repeat it as many times as you want. $\endgroup$ – Mołot Apr 10 '16 at 20:26
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    $\begingroup$ I'd hardly call that "major"! Now I can also draw the image in blue in or black ink or red ink or a pencil and I could draw it upside down or to the left orientation. I can draw it after drinking a glass of milk or after eating dim sum.... $\endgroup$ – fleablood Apr 11 '16 at 17:56
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    $\begingroup$ @fleablood You could draw it in a boat, you could draw it with a goat? You could draw it here or there, you could draw it anywhere? $\endgroup$ – neminem Apr 11 '16 at 23:09
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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. $8\cdot6\cdot4\cdot2\cdot4\cdot2 = 3072$

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    $\begingroup$ This answer is more intuitive than the @ross's answer, if you fleshed out why it's 8 ways, then 6, etc... this would be a great answer. $\endgroup$ – user138559 Apr 10 '16 at 22:16
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    $\begingroup$ I think part of the elegance of this answer is that it doesn't require fleshing out. The answer self-fleshes. There is a hidden elegance even in the meaning of the "That's obvious" comment at the beginning. It sound's dismissive. But it's actually descriptive. $\endgroup$ – Mowzer Apr 11 '16 at 18:43
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    $\begingroup$ Something rubs me the wrong way about "That's obvious". I don't think that language has a place on a Q&A site. If it was obvious, they wouldn't have asked. $\endgroup$ – JPhi1618 Apr 12 '16 at 14:19
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    $\begingroup$ Telling the question asker "That's obvious" violates the Be nice principle that the Stack Exchange network and community tries to adhere to. You can write an answer without belittling the question asker. If it was obvious, then the question wouldn't be asked. Furthermore, you should flesh out what you mean by your various numbers and how you arrived at them. You just stated "8, then 6, then 4..." which is hard to read and understand. And Mowzer, you should not be encouraging this kind of language or vague answers. $\endgroup$ – The Anathema Apr 12 '16 at 17:58
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    $\begingroup$ @TheAnathema I agree it took me a minute and to look back at the picture to see why $8,6\ldots$ make sense. I can imagine it'd take longer if you don't 'see it' and the answer doesn't help you arrive at it, except by giving the numbers. I do however agree it is intuitive once you see it. $\endgroup$ – snulty Apr 13 '16 at 12:15
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Looking at the picture, there are 4 phases.

  1. Draw the petals
  2. Draw the upper stem
  3. Draw the leaves
  4. 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

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

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    $\begingroup$ since the question requires you to draw the petals first, it would be more intuitive to enumerate the options for drawing the petals before enumerating the options for drawing the leaves. $\endgroup$ – Level River St Apr 12 '16 at 18:29
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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?

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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 $z$.

enter image description here

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 enter image description here

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Enumerate the possibilities:

  1. Draw one of 4 top petals, in one of two directions: 4*2
  2. Draw one of 3 remaining top petals, in one of two directions: 3*2
  3. Draw one of 2 remaining top petals, in one of two directions: 2*2
  4. Draw last petal in one of two directions: 2
  5. Draw one of 2 leaves, in one of two directions: 2*2
  6. 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

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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.

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    $\begingroup$ o.0 My eyes... also, doing nothing but citing other users' answers is not a valid answer. See How to Answer $\endgroup$ – cat Apr 13 '16 at 1:08
  • $\begingroup$ I added an explanation. Is that nothing? $\endgroup$ – C Marius Apr 13 '16 at 10:19
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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.

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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. ;)

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    $\begingroup$ It says: "starting from the center of the petals". $\endgroup$ – Mathematician 42 Apr 13 '16 at 13:35
  • $\begingroup$ True, but one could argue that there's nothing in the question actually forcing you to have to complete all the petals first before proceeding to the rest of the flower. Obviously we all assume that you can't retrace your steps beyond crossing a point, but the question doesn't actually specify that, and it really should have. Otherwise I could draw two petals go down and draw one leaf then go back up and draw two more petals and back down to draw the second leaf prior to finishing the stem. Or draw one petal, go down draw one leaf, go up and draw another petal, down to finish the stem, up..... $\endgroup$ – Monomeeth Apr 14 '16 at 3:45

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