Question

Five-digit Numbers How many five-digit numbers can be

formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 if the first

digit cannot be 0 or 1? Repeated digits are allowed.

Short answer

There are 80,000 five-digit numbers can be

formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 if the first digit cannot be 0 or 1.

Step-by-step solution

  Step 1. Given information

$$\begin{gathered} \mathrm{Five}-\mathrm{digit}\mathrm{Number}:\mathrm{abcde} \\ \mathrm{Choices}:0,1,2,3,4,5,6,7,8,\mathrm{and}9 \\ \mathrm{Conditions}:1.\mathrm{First}\mathrm{digit}\mathrm{cannot}\mathrm{be}0\mathrm{or}1. \\ 2.\mathrm{Repeated}\mathrm{digits}\mathrm{are}\mathrm{allowed}. \end{gathered}$$

Step 2. Calculation

$$\begin{gathered} \mathrm{Five}-\mathrm{digit}\mathrm{Number} \\ \mathrm{Lets}\mathrm{say}:\mathrm{abcde} \\ \mathrm{For}\mathrm{a}:8\mathrm{choices} \\ \mathrm{For}\mathrm{b}:10\mathrm{choices} \\ \mathrm{For}\mathrm{c}:10\mathrm{choices} \\ \mathrm{For}\mathrm{d}:10\mathrm{choices} \\ \mathrm{For}\mathrm{e}:10\mathrm{choices} \\ \mathrm{So}, \\ \mathrm{five}-\mathrm{digit}\mathrm{numbers}\mathrm{can}\mathrm{be}\mathrm{formed}\mathrm{using}\mathrm{the}\mathrm{given}\mathrm{digits}: \\ 8\times 10\times 10\times 10\times 10=80,000 \end{gathered}$$