Question

A shopkeeper has $10$ copies each of nine different books, then the number of ways in which atleast one book can be selected is

• A. $9^{11}-1$
• B. $10^{10}-1$
• C. $11^9-1$
• D. $10^9$

Answer 1

Answer: (C)

$\underset{10}{\underbrace{B_1}}\underset{10}{\underbrace{B_2}}\underset{10}{\underbrace{B_3}}\dots \dots ..\underset{10}{\underbrace{B_9}}$
$1^{st}$ kind of book can be selected in $\left(10+1\right)$ ways.
Similarly each kind of book can be selected in $\left(10+1\right)$ ways.
So, selection of atleast one book $=\left(10+1\right)\left(10+1\right)\dots \dots \dots ..\left(10+1\right)\left(9times\right)-1$
$=11^9-1$