Find the number of integral solutions of equation `x+y+z+t=29,` when `x >= 1, y >= 2, z>=3 and t >= 0`

Explanation

Since, x+y+z+t=29 . .. (i)
and x,y,z,t are integer
`therefore x ge 1, y ge 2, z ge 3, t ge 0`
`implies x -1 ge 0, y-2ge0,z-3ge0,tge0`
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)ge0,x_(2)ge0,x_(3)ge0,tge0`
From eq. (i) we get
`x_(1)+1+x_(2)+2+x_(3)+3+t=29`
`impliesx_(1)+x_(2)+x_(3)+t=23`
Hence, total number of solutions`=.^(23+4-1)C_(4-1)`
`=.^(26)C_(3)=(26*25*24)/(1*2*3)=2600`