Question 1202852
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In a certain city, there are seven streets going north-south and five streets going east-west. 
How many street paths start at the southwest corner of the city, end at the northeast corner of the city, 
and have the shortest possible length?
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<pre>
Each such path is a combination of elementary 7 south->north parts/blocs and 
5 west->east parts/blocks, 7+5 = 12 blocks, in total.


So, using designation/letter V for south->north blocks ("vertical") and designation/letter H ("horizontal") 
for west->east blocks, each path can be coded as a word of 12 letters "V" and "H".


There is one-to-one correspondence between such paths and such words.


So, we can say that the number of paths equals the number of words of the length 12,
written using letter V and H.  Repeating of letters is allowed.


From Combinatorics, it is known that the number of such words is  

    {{{12!/(7!*5!)}}} = after reduction of the fraction = {{{(12*11*10*9*8)/(1*2*3*4*5)}}} = {{{C[12]^5}}} = {{{C[12]^7}}} = 792.


Indeed, from the total 12 positions of the word, 7 positions for letters "V" can be selected 
in  {{{C[12]^7}}}  different ways; the rest 5 positions are occupied by the letters "H".


It gives us the answer to the problem's question: there are 792 such paths.
</pre>

Solved.


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The closest combinatorial problem is this:


<pre>
    How many distinguishable ways are there to place in line 7 red and 5 white balls, if they distinct only by color ?"
</pre>


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To see many other similar &nbsp;(and different) &nbsp;solved problems, &nbsp;look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/Arranging-elements-of-sets-containing-undistinguishable-elements.lesson>Arranging elements of sets containing indistinguishable elements</A> 

in this site.