Question

How many integral solutions are there for the equation $x+y+z+w=29$ when $x>0,y>1,z>2$ and $w>0$

• A. $2500$
• B. $4200$
• C. $2300$
• D. none of these

Answer 1

Answer: (C)

$2300$

Given $x+y+z+w=29$ $,$ $w,x>0,y>1,z>2$

Let $w=a,$ $x=b,$ $y-1=c,$ $z-2=d$

Now the equation changes to -

$\Rightarrow a+b+(c+1)+(d+z)=29$ $,$ $a,b,c,d>0$

$\Rightarrow a+b+c+d=26$

Total integral solutions is given by ${}^{n-1}{C}_{r-1}$ Here, $n=26,r=4$

$\Rightarrow {}^{26-1}{C}_{4-1}={}^{25}{C}_3$

Total $=\frac{25\times 24\times 23}{3\times 2}$

$=2300$

Hence, the answer is $2300.$