Question

No. of different ways by which $3$ persons $A,B$ and $C$ having $6$ one rupee coins, $7$ one rupee coins and $8$ one rupee coins respectively can donate $10$ one rupee coins collectively.

• A. $41$
• B. $43$
• C. $47$
• D. $49$

Answer 1

The correct option is A $41$

We know that, the number of non-negative integral solutions to

$x_1+x_2+...+x_n=r$ is $\left({}^{n+r-1}{C}_r\right)$

Hence for $A+B+C=10$

Total number of ways $={}^{10+3-1}{C}_{10}={}^{12}{C}_{10}$

Now, we will have to subtract the cases where $A>6,B>7,C>8$

$Case-I:$

$A>6,B>0,C>0\Rightarrow$ Give $A=7$ already and then find the total non-negative integral solution.

$\therefore A^{\prime }+B+C=10$

Bur, $A^{\prime }=A+7$

$\Rightarrow A+B+C+7=10\Rightarrow A+B+C=3$

$\therefore$ Total no. of ways $={}^{3+3-1}{C}_3={}^5{C}_3$

$Case-II:$

$A>0,B>7,C>0$ ; Similarly as above

$\Rightarrow A+B+C=2$

$\Rightarrow$ Total no. of ways $={}^{3+2-1}{C}_2={}^4{C}_2$

$Case-III:$

$A>0,B>0,C>8$

$\Rightarrow A+B+C=1$

$\Rightarrow$ Total no. of ways $={}^{3+1-1}{C}_1={}^3{C}_1$

$\therefore$ Total number of ways ${=}^{12}{C}_{10}-({}^5{C}_3+{}^4{C}_2+{}^3{C}_1)$

Hence, correct answer is $41$