algebra
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Determine the number of positive integer solutions to the equation x+y+z+t=7
- Fahim Shahriar
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- Joined:Sun Dec 18, 2011 12:53 pm
Re: algebra
When $x=1$ and $y=1$ , $z+t=5$ has $4$ solutions. Keeping $x=1$ fixed, $+1$ the value of $y$ gradually. Then $z+t$ will have $3,2$ and $1$ solutions for $y=2,3,4$ respectively.
So keeping $x=1$ fixed, we get $4+3+2+1=10$ solutions.
Keeping $x=2$ fixed, we get $(10-4)=6$ solutions. [4 solutions out. Because then z+t=5 is not possible].
Similarly, keeping $x=3,x=4$ fixed, we get $(6-3)=3$ and $(3-2)=1$ solutions respectively.
Total $= 10+6+3+1 = \boxed {20}$ $\text{solutions}$.
So keeping $x=1$ fixed, we get $4+3+2+1=10$ solutions.
Keeping $x=2$ fixed, we get $(10-4)=6$ solutions. [4 solutions out. Because then z+t=5 is not possible].
Similarly, keeping $x=3,x=4$ fixed, we get $(6-3)=3$ and $(3-2)=1$ solutions respectively.
Total $= 10+6+3+1 = \boxed {20}$ $\text{solutions}$.
Last edited by *Mahi* on Fri Jan 24, 2014 11:09 am, edited 1 time in total.
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Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
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Re: algebra
how do you understand that $z+t=5$ has 4 solutions?
- Phlembac Adib Hasan
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Re: algebra
$1+4,2+3,3+2,4+1$. In fact $z+t=n$ has $n-1$ solutions in $\mathbb N$.Ridwan Abrar wrote:how do you understand that z+t=5 has 4 solutions?
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