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I'm prepping for the GRE.

Would appreciate if someone could explain the right way to solve this problem. It seems simple to me but the site where I found this problem says I'm wrong but doesn't explain their answer. So here is the problem verbatim:

enter image description here

Find the number of paths from x to y moving only right (R) or down (D).

My answer is 6. What am I missing??

Thanks for any help.

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Well I also count them to be 6. – randomname Jan 12 '14 at 19:52
Your answer is correct. The answer is $6$. – NasuSama Jan 12 '14 at 19:53
OK thanks. This site's test must have made a mistake. It says 28 is the answer. Thanks for the sanity check. – user120865 Jan 12 '14 at 19:58
This site assumes that you need to use the following formula for counting the number of paths, which is $\dbinom{6 + 2}{2}$ or equivalently $\dbinom{6 + 2}{6}$ – NasuSama Jan 12 '14 at 19:58
You do not "walk" on the squares, you walk on the lines. There are $7\times 3$ lines! – Jeppe Stig Nielsen Jan 12 '14 at 21:06

The solution to the general problem is if you must take $X$ right steps, and $Y$ down steps then the number of routes is simply the ways of choosing where to take the down (or right) steps. i.e.

$$ \binom{X + Y}{X} = \binom{X + Y}{Y} $$

So in your example if you are traversing squares then there are 5 right steps and 1 down step so:

$$ \binom{6}{1} = \binom{6}{5} = 6 $$

If you are traversing edges then there are 6 right steps and 2 down steps so:

$$ \binom{8}{2} = \binom{8}{6} = 28 $$

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It may be worth noting for the OP that a similar method works if $x,y$ are in $n$-dimensional space. – Alyosha Jan 17 '14 at 17:49

Any such path is a permutation of 6 R and 2 D, so the answer is $${6+2\choose 2}={8\choose 2}=\frac{{8 \times 7}}{2}=28$$

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maybe this is the answer traveling along edges – janmarqz Jan 12 '14 at 20:03
there goes my +1 'cuz this has to do with the Catalan numbers :D – janmarqz Jan 12 '14 at 20:05
Exactly janmarqz! This rectangle is called a 7 by 3 grid, so the traveling is along vertices on line segments. – Woria Jan 12 '14 at 20:07
I agree, this could have been clearer, but the $x$ and $y$ seem to mark lattice points, not squares. – Carsten S Jan 12 '14 at 20:17

If you panic during the test, consider just drawing it as a Pascal's triangle


  • Value at the origin (x) is 1
  • For all other nodes, the value is the sum of its top and left neighbors

Pascal's triangle

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interesting thanks! – Grijesh Chauhan Jan 13 '14 at 10:11

To get from point x (not square x) to point y there are $8$ steps to be taken. $2$ of them downwards and $6$ to the right. So it just comes to electing exactly $2$ of the $8$ consecutive steps to be the steps downwards.

Picking $2$ out of $8$ can be done on $\binom{8}{2}=28$ ways.

When you are thinking in squares instead of points then there $6$ steps to be taken. $1$ of them downwards and $5$ to the right. So it just comes to electing exactly $1$ of the $6$ consecutive step to be the steps downwards.

Picking $1$ out of $6$ can be done on $\binom{6}{1}=6$ ways.

That explains the fact that your answer was $6$.

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So the lesson is to look carefully at what is being asked. In this case movement between vertices rather then hops between squares, and vertices because you have to assume based on the diagram that X and Y refer to vertices not squares. Seems like an easy mistake to make. – user120865 Jan 12 '14 at 20:30
Indeed. A good look at what is being asked is beyond doubt always the best start to solve the problem. – drhab Jan 12 '14 at 20:37

See here you can easily compute the number of paths by applyinfg the COMBINATION concept.

So what is the concept!!

ok then.....let us consider to proceed from x to y we have to take the steps to the direction of RIGHT and DOWN....fine?

and let us take RIGHT = R and DOWN = D.....still fine.. :-)

now see if anyone wants to go from X to Y he must take 2 DOWN steps and atmost 6 RIGHT step.

So we can say that we have to take total 8 numbers of step.You can check it by your own that we must take 8 steps to reach at Y starting from X poit.ok.

and see here we can write a combination of steps,like: { R-R-R-R-D-R-D-R } -----(see its a suitable traverse path.match this path with the picture.)

And now we can make our main observation on this problem.....So what is it? See, you have to take total of 8 steps and all of them are the combination of R and D's ok?

And the observation is: You have to take just 2 D(DOWN step) and also just 6 R(RIGHT step) to reach at Y.

So we can say that the total number of combination is: HOW MANY COMBINATION OF 'R' OR 'D' is possible from the total of 8 steps!!!!!

The required answer is actually the combination of 2 PLACES for D's OR 6 PLACES for R's from 8 total steps(OR TOTAL 8 PLACES where R and D letters take place)........

see here we are just mapping our problem of counting the number of suitable paths from X to Y into a basic counting problem.....And we just logically made a connection between these two problems and it is clear that if we can calculate the combination of 2 PLACES for D's OR 6 R's from total 8 steps it will be enough!!

Here is the total of 8 places where D and R may take places:

{ ___  ___  ___  ___  ___  ___  ___  ___ }

And we actually CHOOSE 2 places for for 2 D's OR we CHOOSE 6 places for 6 R's but we dont bother for the oreding of D's OR R's.....the 2 D's OR the 6 R's can tale place randomly but their requirement for the places is fixed!! it is 2 for 2 D's and it is 6 for 6 D's......actually choosing 2 D's OR 6 R's are basically same!! :-)

So, the answer is : 8C2 or 8C6 (note that both are same!! :-) ) 8C2 = 28 and easily you can see 8C6 = 28.


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Thanks, great answers. – user120865 Jan 12 '14 at 20:43
you have read the whole answer given by me!nice...and it is my pleasure to help you. – arkadeep Jan 12 '14 at 20:57

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\ot}{\downarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ In general we have steps to right and down. Tipically, this is a possible path: $$ {\Huge \to\ \to\ \down\ \to\ \to\ \to\ \down \to} $$ In general, the problem is reduced to put two down arrows $\down$ in 8 sites like the following example: $$ \begin{array}{cccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ &&&&\down&&\down \end{array} $$ Let's put one $\down$ at the first step, then we have $\color{#ff0000}{\large 7}$ positions left to put the other one. Next, we put the first $\down$ at the second step and we have $\color{#ff0000}{\large 6}$ position left to put the other one and so on such that $7 + 6 + 5 + 4 + 3 + 2 + 1 = \color{#0000ff}{\Large 28}$ which is $\ds{8 \choose 2}$.

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Check this handcrafted portable network graphic : enter image description here

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You walk on the squares, but they want you to walk on the black lines. – Jeppe Stig Nielsen Jan 12 '14 at 21:08
yes, this is the interpretation in the OP. Others are explained aside. – janmarqz Jan 13 '14 at 2:43

First note that you will always be moving 6 squares right and 2 squares down, just because $y$ is 2 squares down and 6 squares to the right from $x$.

Next, note that a different ordering of these steps will always result in a different path. Then, the question simply becomes how many ways there are to arrange 6 "right steps" and 2 "down steps" from first to last.

This is equal to the total number of ways to arrange 8 steps ($= 8!$), divided by the number of different ways that are accounted for by some of the steps being the same ($6!$ for 6 right steps and $2!$ for 2 down steps), which is $\displaystyle \frac{8!}{6!2!} = 28$.

If you're moving only 5 squares right and 1 square down, which is the case if you're moving on the insides of squares rather than the intersections of grid lines, then you're arranging 6 total steps, 5 right and 1 down, which gives the formula $\displaystyle \frac{6!}{5!1!} = 6$.

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