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 $0,1,2,3,4\dots \dots 9$ 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 $=\frac{{}^{10}{C}_4\times 4!}{4!}={}^{10}{C}_4$