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?


1 Answer 1


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 $7,8$.

  • $\begingroup$ the first it is not $24$? $\endgroup$
    – Legolas
    Nov 22, 2015 at 18:55
  • $\begingroup$ @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 ($0,2,4,6,8$). $\endgroup$
    – user236182
    Nov 22, 2015 at 19:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .