Question

We know that the number of positive integral solutions of the equation $x_1+x_2+x_3+...+x_m=n\left(nϵN\right)$ is equal to the coefficient of $x^n$ in the expansion of ${(x+x^2+x^3+...to\infty )}^m$ when $\left|x\right|<1$. Also, we have the expansion ${(1-x)}^{-n}{=}^{n-1}{C}_0{+}^n{C}_{1}x{+}^{n+1}{C}_2{x}^2+...{+}^{n+r}{C}_{r+1}{x}^{r+1}+...to\infty ,$

The number of ways in which 10 identical things can be distributed among 3 persons so that each gets at least one thing is

• A. ${}^8{C}_3$
• B. ${}^{10}{C}_4$
• C. ${}^9{C}_3$
• D. ${}^9{C}_2$

Answer 1

The correct option is D ${}^9{C}_2$

Let A,B,C be three people.
Hence considering A gets only 1 object, then the remaining 9 object will be distributed among the rest 2, in 8 different way
Considering A has 2, there will be 7 ways to distribute the things among rest 2.. and so on.
Hence the required permutation will be
$8+7+6+5+..1$
$=\frac{8(8+1)}{2}$
$=\frac{9\left(8\right)}{2}$
$=\frac{9(9-1)}{2}$
$={}^9{C}_2$