Question

The number of integral solutions of equation $x+y+z+t=29$, when $x\ge 1,y\ge 2,z\ge 3$ and $t\ge 0$ is

• A. 2600
• B. 2500
• C. 2550
• D. 2700

Answer 1

Answer: (A)

2600

PROBLEMS BASED ON CERTAIN THEOREMS ON COMBINATIONS :

$\cdot$ The number of positive integral solutions of the equation $x_1+x_2+x_3+....+x_r=n$ is ${}^{n-1}{C}_{r-1}.$$\cdot$ The number of non-negative integral solutions of the equation<br>$x_1+x_2+x_3+....+x_r=n$ is ${}^{n+r-1}{C}_{r-1}.$

$x+y+z+t=29$
Let $x^{\prime }=x-1,y^{\prime }=y-2,z^{\prime }=z-3,t^{\prime }=t$ where $x^{\prime },y^{\prime },z^{\prime },t^{\prime }\ge 0$
$\Rightarrow x^{\prime }+y^{\prime }+z^{\prime }+t^{\prime }=29-1-2-3=23$
here $n=23,r=4$
Thus required number of ways is $={}^{n+r-1}{C}_{r-1}{=}^{26}{C}_3=2600$