Question

The number of integral solutions of equation \[x+y+z+t=\text{ }29\], when \[x\ge 1,\,y\ge 2,\,z\ge 3,\,\text{and}\,t\ge 0\] is , choose the correct answer.
A . 2600
B . 2500
C . 2550
D . 2700

Answer 1

Hint: Assume that there are 4 persons namely x, y,z and t. Now, you are distributing 29 things to them with the condition that \[x\ge 1,\,y\ge 2,\,z\ge 3,\,\text{and}\,t\ge 0\]. So here, x will get minimum 1 thing, y will get minimum 2 things, z will get minimum 3 things and finally t will get minimum 0 things. So, after distributing, we are left with 23 objects that can be distributed about there 4 people in \[{}^{n+r-1}{{C}_{r-1}}\]. Where \[n=\text{ }23\]and \[r=4\].

Complete step-by-step answer:
In the question, we have to find the number of integral solutions of equation \[x+y+z+t=\text{ }29\], when \[x\ge 1,\,y\ge 2,\,z\ge 3,\,\text{and}\,t\ge 0\]. So here lets assume that there are 4 persons namely x, y,z and t, to whom we have to distribute 29 things. Now, the condition that is given is \[x\ge 1,\,y\ge 2,\,z\ge 3,\,\text{and}\,t\ge 0\]which means that x will get minimum 1 thing, y will get minimum 2 things, z will get minimum 3 things and finally t will get minimum 0 things. Next, when we have already distributed the minimum number of things then we are left with \[29-\left( 1+2+3+0 \right)=23\]things. So, now we just have to find the number of ways of distributing the remaining 23 things to the 4 people. This is given by the formula \[{}^{n+r-1}{{C}_{r-1}}\], where n is the number of things, which is 23 here, and r is 4 here. So, we have:
\[\begin{align}
  & \Rightarrow {}^{n+r-1}{{C}_{r-1}} \\
 & \Rightarrow {}^{23+4-1}{{C}_{4-1}} \\
 & \Rightarrow {}^{26}{{C}_{3}} \\
 & \Rightarrow \dfrac{26!}{3!(26-3)!}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because {}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!} \\
 & \Rightarrow \dfrac{26!}{3!(23)!} \\
 & \Rightarrow \dfrac{26\times 25\times 24}{3\times 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because 3!=3\times 2\times 1 \\
 & \Rightarrow 2600 \\
\end{align}\]
So, finally we have 2600 integral solution of the given equation \[x+y+z+t=\text{ }29\], when \[x\ge 1,\,y\ge 2,\,z\ge 3,\,\text{and}\,t\ge 0\]. Hence the correct answer is option A.

Note: Here the integral solution will mean that we have to find the integer solution and not the fractional solution. So, we have to neglect the fractional solutions. Here number of ways of distributing r things from n things will be \[{}^{n+r-1}{{C}_{r-1}}\]and not \[{}^{n}{{C}_{r}}\].