Mathematics - Permutation & Combination Question with Solution | TestHub

Total number of arrangements of 10 digits $0,1,2,\dots ,9$ 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. $$=\frac{{}^{10}{\mathrm{C}}_4\times 4!}{4!}={}^{10}{\mathrm{C}}_4$$