Number of ways to go to goal vertex from start vertex












1












$begingroup$


I want to find out the number of ways to reach out to goal vertex from start vertex. The problem scenario is depicted as in the picture :



enter image description here



Here I want to go G vertex from S vertex. The only condition here is I can go up and right direction. DIrection diagonally/left/down is not allowed.



So keeping this condition in mind, what is the number of ways to go G vertex from S vertex?



My trying :



I have figured out the no of ways are, $1^2+2^2+3^3+ldots+(n-2)^2$. Is this correct?










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    can you explain how did you get the sum of squares formula?
    $endgroup$
    – Siong Thye Goh
    Dec 31 '18 at 15:46










  • $begingroup$
    hint: how many steps you need to go up (U), right (R)? combinations between them...UURURR, UUURRR....
    $endgroup$
    – dmtri
    Dec 31 '18 at 15:51


















1












$begingroup$


I want to find out the number of ways to reach out to goal vertex from start vertex. The problem scenario is depicted as in the picture :



enter image description here



Here I want to go G vertex from S vertex. The only condition here is I can go up and right direction. DIrection diagonally/left/down is not allowed.



So keeping this condition in mind, what is the number of ways to go G vertex from S vertex?



My trying :



I have figured out the no of ways are, $1^2+2^2+3^3+ldots+(n-2)^2$. Is this correct?










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    can you explain how did you get the sum of squares formula?
    $endgroup$
    – Siong Thye Goh
    Dec 31 '18 at 15:46










  • $begingroup$
    hint: how many steps you need to go up (U), right (R)? combinations between them...UURURR, UUURRR....
    $endgroup$
    – dmtri
    Dec 31 '18 at 15:51
















1












1








1





$begingroup$


I want to find out the number of ways to reach out to goal vertex from start vertex. The problem scenario is depicted as in the picture :



enter image description here



Here I want to go G vertex from S vertex. The only condition here is I can go up and right direction. DIrection diagonally/left/down is not allowed.



So keeping this condition in mind, what is the number of ways to go G vertex from S vertex?



My trying :



I have figured out the no of ways are, $1^2+2^2+3^3+ldots+(n-2)^2$. Is this correct?










share|cite|improve this question











$endgroup$




I want to find out the number of ways to reach out to goal vertex from start vertex. The problem scenario is depicted as in the picture :



enter image description here



Here I want to go G vertex from S vertex. The only condition here is I can go up and right direction. DIrection diagonally/left/down is not allowed.



So keeping this condition in mind, what is the number of ways to go G vertex from S vertex?



My trying :



I have figured out the no of ways are, $1^2+2^2+3^3+ldots+(n-2)^2$. Is this correct?







sequences-and-series number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 15:50









saulspatz

17k31435




17k31435










asked Dec 31 '18 at 15:44









Standard EquationStandard Equation

247113




247113








  • 2




    $begingroup$
    can you explain how did you get the sum of squares formula?
    $endgroup$
    – Siong Thye Goh
    Dec 31 '18 at 15:46










  • $begingroup$
    hint: how many steps you need to go up (U), right (R)? combinations between them...UURURR, UUURRR....
    $endgroup$
    – dmtri
    Dec 31 '18 at 15:51
















  • 2




    $begingroup$
    can you explain how did you get the sum of squares formula?
    $endgroup$
    – Siong Thye Goh
    Dec 31 '18 at 15:46










  • $begingroup$
    hint: how many steps you need to go up (U), right (R)? combinations between them...UURURR, UUURRR....
    $endgroup$
    – dmtri
    Dec 31 '18 at 15:51










2




2




$begingroup$
can you explain how did you get the sum of squares formula?
$endgroup$
– Siong Thye Goh
Dec 31 '18 at 15:46




$begingroup$
can you explain how did you get the sum of squares formula?
$endgroup$
– Siong Thye Goh
Dec 31 '18 at 15:46












$begingroup$
hint: how many steps you need to go up (U), right (R)? combinations between them...UURURR, UUURRR....
$endgroup$
– dmtri
Dec 31 '18 at 15:51






$begingroup$
hint: how many steps you need to go up (U), right (R)? combinations between them...UURURR, UUURRR....
$endgroup$
– dmtri
Dec 31 '18 at 15:51












1 Answer
1






active

oldest

votes


















0












$begingroup$

In your case there is a direct counting approach. Your path is a sequence of $n$ right steps (R's) and $m$ up steps (U's), with $m=n=3$. So you are asking how many strings can be made from 3 R's and 3 U's, which should be $binom{6}{3}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So for n = 4 , the no of ways are $ binom{8}{4} $ = 70 ?
    $endgroup$
    – Standard Equation
    Dec 31 '18 at 15:53








  • 1




    $begingroup$
    Isn't it $n-1$ R's and $m-1$ U's?
    $endgroup$
    – saulspatz
    Dec 31 '18 at 16:29










  • $begingroup$
    @saulspatz thanks fixed
    $endgroup$
    – gt6989b
    Dec 31 '18 at 16:56











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









0












$begingroup$

In your case there is a direct counting approach. Your path is a sequence of $n$ right steps (R's) and $m$ up steps (U's), with $m=n=3$. So you are asking how many strings can be made from 3 R's and 3 U's, which should be $binom{6}{3}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So for n = 4 , the no of ways are $ binom{8}{4} $ = 70 ?
    $endgroup$
    – Standard Equation
    Dec 31 '18 at 15:53








  • 1




    $begingroup$
    Isn't it $n-1$ R's and $m-1$ U's?
    $endgroup$
    – saulspatz
    Dec 31 '18 at 16:29










  • $begingroup$
    @saulspatz thanks fixed
    $endgroup$
    – gt6989b
    Dec 31 '18 at 16:56
















0












$begingroup$

In your case there is a direct counting approach. Your path is a sequence of $n$ right steps (R's) and $m$ up steps (U's), with $m=n=3$. So you are asking how many strings can be made from 3 R's and 3 U's, which should be $binom{6}{3}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So for n = 4 , the no of ways are $ binom{8}{4} $ = 70 ?
    $endgroup$
    – Standard Equation
    Dec 31 '18 at 15:53








  • 1




    $begingroup$
    Isn't it $n-1$ R's and $m-1$ U's?
    $endgroup$
    – saulspatz
    Dec 31 '18 at 16:29










  • $begingroup$
    @saulspatz thanks fixed
    $endgroup$
    – gt6989b
    Dec 31 '18 at 16:56














0












0








0





$begingroup$

In your case there is a direct counting approach. Your path is a sequence of $n$ right steps (R's) and $m$ up steps (U's), with $m=n=3$. So you are asking how many strings can be made from 3 R's and 3 U's, which should be $binom{6}{3}$.






share|cite|improve this answer











$endgroup$



In your case there is a direct counting approach. Your path is a sequence of $n$ right steps (R's) and $m$ up steps (U's), with $m=n=3$. So you are asking how many strings can be made from 3 R's and 3 U's, which should be $binom{6}{3}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 31 '18 at 16:55

























answered Dec 31 '18 at 15:50









gt6989bgt6989b

35.1k22557




35.1k22557












  • $begingroup$
    So for n = 4 , the no of ways are $ binom{8}{4} $ = 70 ?
    $endgroup$
    – Standard Equation
    Dec 31 '18 at 15:53








  • 1




    $begingroup$
    Isn't it $n-1$ R's and $m-1$ U's?
    $endgroup$
    – saulspatz
    Dec 31 '18 at 16:29










  • $begingroup$
    @saulspatz thanks fixed
    $endgroup$
    – gt6989b
    Dec 31 '18 at 16:56


















  • $begingroup$
    So for n = 4 , the no of ways are $ binom{8}{4} $ = 70 ?
    $endgroup$
    – Standard Equation
    Dec 31 '18 at 15:53








  • 1




    $begingroup$
    Isn't it $n-1$ R's and $m-1$ U's?
    $endgroup$
    – saulspatz
    Dec 31 '18 at 16:29










  • $begingroup$
    @saulspatz thanks fixed
    $endgroup$
    – gt6989b
    Dec 31 '18 at 16:56
















$begingroup$
So for n = 4 , the no of ways are $ binom{8}{4} $ = 70 ?
$endgroup$
– Standard Equation
Dec 31 '18 at 15:53






$begingroup$
So for n = 4 , the no of ways are $ binom{8}{4} $ = 70 ?
$endgroup$
– Standard Equation
Dec 31 '18 at 15:53






1




1




$begingroup$
Isn't it $n-1$ R's and $m-1$ U's?
$endgroup$
– saulspatz
Dec 31 '18 at 16:29




$begingroup$
Isn't it $n-1$ R's and $m-1$ U's?
$endgroup$
– saulspatz
Dec 31 '18 at 16:29












$begingroup$
@saulspatz thanks fixed
$endgroup$
– gt6989b
Dec 31 '18 at 16:56




$begingroup$
@saulspatz thanks fixed
$endgroup$
– gt6989b
Dec 31 '18 at 16:56


















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