Question

The number of integral solution of x+y+z+t=29, when $x\ge 1,y>1,z>2,t\ge o$

Answer 1

$\begin{array}{c}Given,\\ x+y+z+t=\mathrm{29.......}whenx\ge 1,y\ge 2,z\ge 3andt\ge 0\\ thetotaldistinctobject=29,distributeintox,y,z,andt\\ x-----1\\ y-----2\\ z-----3\\ then,remaning=29-6\\ =23\\ Now,\\ No.ofways=n+r-1_{C_{r-1}}\\ withoutanyconditiondistributeintox,y,z,t\\ =23+4-1_{C_{4-1}}\\ ={26}_{C_3}\\ =\frac{|!26}{|!3|!23}\\ =\frac{26\times 25\times 24\times |!23}{3\times 2\times 1\times |!23}\\ =2600\end{array}$

so, the Answer is $2600$.