Question

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

• A. $^{10}C_4 \times 4!$
• B. $^{10}C_4$
• C. $\frac{10!}{4!}$
• D. None of these

Answer 1

Answer: (B)

$^{10}C_4$

Answer 2

Total number of arrangements of $10$ digits $0,1,2, \ldots ., 9$
by taking 4 at a 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.
$=\frac{{ }^{10} C_{4} \times 4 !}{4 !}={ }^{10} C_{4}$