Question

The total number of 4-digit numbers in which the digits are descending order is

• A. ${{}^{10}C}_4\times 4!$
• B. ${{}^{10}C}_4$
• C. $\frac{10!}{4!}$
• D. None of these

Answer 1

The correct option is B ${{}^{10}C}_4$

Total number of arrangements of $10$ digits $0,1,2,....9$ by taking $4$ at at time ${{}^{10}C}_4\times 4!$
We observe that in every arrangement of $4$ selected digits there is just one arrangement in which the digits are in descending order
$\therefore$ Required number of $4$-digit numbers
$=\frac{{{}^{10}C}_4\times 4!}{4!}={{}^{10}C}_4$