Question

In how many ways can Rs.$16$ be divided into $4$ persons when none of them get less than Rs.$3$ ?

• A. $70$
• B. $35$
• C. $64$
• D. $190$

Answer 1

The correct option is B

$35$

Explanation for the correct option:

Finding the number of ways Rs.$16$ can be divided into $4$ persons when none of them gets less than Rs.$3$:

Defining $x_{1,}{x}_{2,}{x}_{3,}{x}_4$as variables such that

$x_1+x_2+x_3+x_4=16.....\left(i\right)$

According to the given information $x_1\ge 3,x_2\ge 3,x_3\ge 3,x_4\ge 3$

Again defining $X_{1,}{X}_{2,}{X}_{3,}{X}_4$ all greater than $0$.

Then we can write

$$\begin{gathered} x_1=3+X_1 \\ {x}_2=3+X_2 \\ {x}_3=3+X_3 \\ {x}_4=3+X_4 \end{gathered}$$

Substituting the above values into $\left(i\right)$

$\begin{array}{rcl}& \Rightarrow & X_1+X_2+X_3+X_4+12=16\\ & \Rightarrow & X_1+X_2+X_3+X_4=4.......\left(ii\right)\end{array}$

We know that number of solutions for $X_1+X_2+X_3+X_4+........+X_k=n$ = ${}^{n+k-1}{C}_{k-1}$

From equation $\left(ii\right)$ we have $k=4\&n=4$

Thus the required number of ways to dividing Rs. $16$ accordingly $={}^{7}C_3$

$$\begin{gathered} =\frac{7!}{3!\times (7-3)!} \\ =35 \end{gathered}$$

Hence, option (B) is the correct answer.