The total number of 4 -digit numbers in which the digits are in...

Total number of arrangements of 10 digits $0,1,2, \ldots, 9$ bytaking 4 at atime $={ }^{10} \mathrm{C}_{4} \times 4 !$ 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. =$\frac{{ }^{10} \mathrm{C}_{4} \times 4 !}{4 !}={ }^{10} \mathrm{C}_{4}$