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How many distinct solutions does the following equation have?

$$x _ { 1 } + x _ { 2 } + x _ { 3 } + x _ { 4 } = 100 $$$x _ { 1 } \in \{ 1,2,3 . . \} , x _ { 2 } \in \{ 2,3,4 , \ldots \} , x _ { 3 } , x _ { 4 } \in \{ 0,1,2,3 , \dots \} $

2 Answers

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x1 + x2 + x3 + x4 = 100

x1 >= 1 ,x2 >= 2 , x3 >= 0 , x4>= 0

1<= x1 <= 98 ( x2 = 2 atleast)

2<= x2 <= 99

0 <= x3,x4 <= 97

the required number of solution = coefficient of x100    in  (x + x2 + x3 + --  x98) (x2 + x3 + x4 +--- + x99)(x0+ x1 + x2 + --- + x97)(x0+ x1 + x2 + --- + x97)

= coefficient of x100    in x(1 + x + x2 + -- +x97) x2 (1 + x + x2 + -- + x97)(1 + x + x2 +-- +x97)(1 + x + x2 +-- +x97)

= coefficient of x100    in x3  (1 + x + x2 + -- +x97) (1 + x + x2 + -- + x97)(1 + x + x2 +-- +x97)(1 + x + x2 +-- +x97)

= coefficient of x97   in  (1 + x + x2 + -- +x97) 4

= coefficient of x97   in (( 1-x98)/(1-x))4

=  coefficient of x97   in (1 -x98 ) 4 (1-x)-4

 = coefficient of x97   in (1 - x) -4   

= 97 + 4 - 1 C 97  

= 161700

edited by
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@Raja Raval solved it Very nicely. I am trying one more approach here.

We have to find

$$x _ { 1 } + x _ { 2 } + \ldots + x _ { k } = n , \text { where } x _ { i } \in \{ 0,1,2,3 , \dots \}$$

                                           if all x1,x2,..xn belongs to {0,1,2,3....}

     answer will be            $\left( \begin{array} { c } { n + k - 1 } \\ { k -1 } \end{array} \right) = \left( \begin{array} { c } { 100 + 4  - 1 } \\ { 4 - 1 } \end{array} \right)$

why because it is same as filling n holes  from k type of bagels replacement  (Watch Above Video !!!)

but our x1 >= 1 and x2 >= 2  so if we write y1 = x1 - 1 and y2 = x2 - 2

then we are left with n = 100 - (1+2) = 97 ( why because .... we have already chosen 1 object of x1 and 2 objects of x2)

our problem boils down to

$$y _ { 1 } + y_ { 2 } + \ldots + x _ { k } = n , \text { where } x _ { i } \in \{ 0,1,2,3 , \dots \}$$

                                            all y1,y2,x2..xk belongs to {0,1,2,3....}

So the Answer is

                                                 $$ \left( \begin{array} { c } { n + k - 1 } \\ { k -1 } \end{array} \right) = \left( \begin{array} { c } { 97 + 4  - 1 } \\ { 4 - 1 } \end{array} \right) = \left( \begin{array} { c } { 100 } \\ { 3 } \end{array} \right) = \left( \begin{array} { c } { 100 } \\ { 97 } \end{array} \right) $$

 

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