Question

How many integral solutions are there to $x+y+z+t=29$, when $x\ge 1,y\ge 1,z\ge 3$ and $t\ge 0$?

Answer 1

Since $x+y+z+t=29...\left(i\right)$
and $x,y,z,t$ are integers
$\therefore x\ge 1,y\ge 2,z\ge 3,t\ge 0$
$\Rightarrow x-1\ge 0,y-2\ge 0,z-3\ge 0,t\ge 0$
Let $x_1=x-1,x_2=y-2,x_3=z-3$
Or $x=x_1+1,y=x_2+2,z=x_3+3$ and then $x_1\ge 0,x_2\ge 0,x_3\ge 0,t\ge 0$
From $\left(1\right)$,
$x_1+1+x_2+1x_3+1+t=29$
$\Rightarrow x_1+x_2+x_3+t=23$
Hence total number of solutions
$={}^{23+4-1}{C}_{4-1}$
$={}^{26}{C}_3=\frac{\mathrm{26.25.24}}{\mathrm{1.2.3}}=2600$