# Number of positive integral solutions to equation x+y+z+w=10 (x>=0,y>=0,z>=2,w>=2) isPlZZZZZZZZZZ answer soon.

jagdish singh singh
173 Points
7 years ago
$\hspace{-0.70 cm}Let z-2=t\geq 0 and w-2=u\geq 0 So eqn.convert into\\\\ x+y+t+2+u+2=10\Rightarrow x+y+z+u=8\;, Where x,y,z,u\geq0\\\\Which is equivalent to arrange 8 stars and 3 bars in row in that way.\\\\\underbrace{*}_{x}|\underbrace{-}_{y}|\underbrace{****}_{z}|\underbrace{***}_{u}, He x=1\;,y=0,z= 4,u=3 is one example\\\\ S Total no.of ways of arranging 8 stars and 3 bars Row is = \binom{8!}{3!\times 5!}$
jagdish singh singh
173 Points
7 years ago
Sorry I have calculate for non negative Integral solution...................................................
jagdish singh singh
173 Points
7 years ago
To Admin when i post my answer it did not take My answer and show a message answer already exists.plz help me. anywhy answer is $\displaystyle \binom{11}{3}$
gandhi1234
10 Points
7 years ago
But the answer given is 84!!!!!!!!!.How????????????????.Plz provide the answer in a complete manner.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

jagdish singh singh
173 Points
7 years ago
Actually You are Right Answer is =84. Here is my solution.......................................................$\hspace{-.70 cm} Let x+1=x'\geq 1\;,y+1=y'\geq 1 and z-1=z'\geq 1\;,u-1=u'\geq 1\\\\So equation is x'+y'+z'+u'=10\;, Where x',y',z',u'\geq 0\\\\Which is equivalent to arange 3 bars in between 10 stars.\\\\ Like this way \underbrace{***}_{x'}|\underbrace{**}_{y'}|\underbrace{****}_{z'}|\underbrace{*}_{u'}\\\\So Total no. of ways is =\binom{9}{3} = 84.$
gandhi1234
10 Points
7 years ago
But z>=2 and w>=2 and you take z>=1 and w>=1.How is it possible?My answer is coming out to be 7C3.Is it correct???????????
The text passage contains a mathematical question and multiple answers provided by different users. The question is about finding the number of positive integral solutions to a given equation with specific constraints.
single_question_with_answer

the question is:
"number of positive integral solutions to equation x+y+z+w=10 (x>=0, y>=0, z>=2, w>=2)"

the correct answer provided by jagdish singh singh is:
\(\binom{9}{3} = 84\)

this is the correct solution, as it accounts for the constraints \(z \geq 2\) and \(w \geq 2\) by transforming the variables and then using the stars and bars method to find the number of solutions.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Number of positive integral solutions to equation x+y+z+w=10 (x>=0,y>=0,z>=2,w>=2) isPlZZZZZZZZZZ answer soon.

jagdish singh singh
173 Points
7 years ago
$\hspace{-0.70 cm}Let z-2=t\geq 0 and w-2=u\geq 0 So eqn.convert into\\\\ x+y+t+2+u+2=10\Rightarrow x+y+z+u=8\;, Where x,y,z,u\geq0\\\\Which is equivalent to arrange 8 stars and 3 bars in row in that way.\\\\\underbrace{*}_{x}|\underbrace{-}_{y}|\underbrace{****}_{z}|\underbrace{***}_{u}, He x=1\;,y=0,z= 4,u=3 is one example\\\\ S Total no.of ways of arranging 8 stars and 3 bars Row is = \binom{8!}{3!\times 5!}$
jagdish singh singh
173 Points
7 years ago
Sorry I have calculate for non negative Integral solution...................................................
jagdish singh singh
173 Points
7 years ago
To Admin when i post my answer it did not take My answer and show a message answer already exists.plz help me. anywhy answer is $\displaystyle \binom{11}{3}$
gandhi1234
10 Points
7 years ago
But the answer given is 84!!!!!!!!!.How????????????????.Plz provide the answer in a complete manner.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

jagdish singh singh
173 Points
7 years ago
Actually You are Right Answer is =84. Here is my solution.......................................................$\hspace{-.70 cm} Let x+1=x'\geq 1\;,y+1=y'\geq 1 and z-1=z'\geq 1\;,u-1=u'\geq 1\\\\So equation is x'+y'+z'+u'=10\;, Where x',y',z',u'\geq 0\\\\Which is equivalent to arange 3 bars in between 10 stars.\\\\ Like this way \underbrace{***}_{x'}|\underbrace{**}_{y'}|\underbrace{****}_{z'}|\underbrace{*}_{u'}\\\\So Total no. of ways is =\binom{9}{3} = 84.$
gandhi1234
10 Points
7 years ago
But z>=2 and w>=2 and you take z>=1 and w>=1.How is it possible?My answer is coming out to be 7C3.Is it correct???????????
user
The text passage contains a mathematical question and multiple answers provided by different users. The question is about finding the number of positive integral solutions to a given equation with specific constraints.

TYPE: SINGLE_QUESTION_WITH_ANSWER

The question is:
"Number of positive integral solutions to equation x+y+z+w=10 (x>=0, y>=0, z>=2, w>=2)"

The correct answer provided by jagdish singh singh is:
\(\binom{9}{3} = 84\)

This is the correct solution, as it accounts for the constraints \(z \geq 2\) and \(w \geq 2\) by transforming the variables and then using the stars and bars method to find the number of solutions.