how many four-digit numbers that have only even digits can be formed?How many three-digit numbers that do not contain the digit $7$ but contain at least one time $8$ can be formed?
Could anyone help me with any hints?
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First question: $4\cdot 5\cdot 5\cdot 5$. Second question: $8\cdot 9\cdot 9-7\cdot 8\cdot 8$, i.e. the number of three-digit numbers without $7$ minus the number of three-digit numbers without both $7,8$.
answered Nov 22, 2015 at 18:42
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$\begingroup$ @Legolas It's $4\cdot 5\cdot 5\cdot 5$. There are $4$ choices for the first digit ($2,4,6,8$), then $5$ choices for each of the other digits ($0,2,4,6,8$). $\endgroup$ Commented Nov 22, 2015 at 19:42