Question

The total number of $4$ digit numbers in which the digits are in ascending order is

• A. ${ }^{10} C _{4}$
• B. $\frac{10 !}{4 !}$
• C. $\frac{{ }^{10} C_{4}}{4 !}$
• D. ${ }^{10} C _{4} \times 4$
• E. $10 ! \times 4 !$

Answer 1

Answer: (A)

The total number of arrangements of $10$ digits $0,1,2,3,4 \ldots \ldots 9$ by taking $4$ at a time
$={ }^{10} C_{4} \times 4 !$
In every arrangement of $4$ selected digits there is just one arrangement in which digits are in ascending order.
Hence, Total $=\frac{{ }^{10} C_{4} \times 4 !}{4 !}={ }^{10} C_{4}$