Question

How many integral solutions are there to $x+y+z+t=29$, when $x\ge 1,y\ge 2,z\ge 3$ and $t\ge 0$?

• A. ${}^{29}C_3$
• B. ${}^{26}C_3$
• C. ${}^{23}C_3$
• D. ${}^{21}C_3$

Answer 1

The correct option is B

${}^{26}C_3$

Explanation for the correct option:

Step 1. Given equation :

$x+y+z+t=29$ where. $x,y,z$and $t$ are integers.

$x\ge 1,y\ge 2,z\ge 3,t\ge 0$

$\Rightarrow (x-1)\ge 0,(y-2)\ge 0,(z-3)\ge 0,t\ge 0$

Step2. find the total number of solutions :

let $x_1=(x-1),x_2=(y-2),x_3=(z-3)$

or $x=x_1+1,y=x_2+2,z=x_3+3$

Putting these values in the given equation. we get

$$\begin{gathered} x_1+1+x_2+2+x_3+3+t=29 \\ \Rightarrow x_1+x_2+x_3+t=23 \end{gathered}$$

Total number of solutions $={}^{23+4-1}C_{4-1}={}^{26}C_3$

$\left[\because numberofsolutionof{x}_1+x_2+....+x_r=kis{}^{k+r-1}C_{r-1}\right]$

Hence the correct option is B.