Discussion
Let \(A = \{0, 2, 4, 6, 8\}\). The number of 5-digit numbers that can be formed using the digits in \(A\) without replacement, is
Answer: B
There are 96 such numbers.
Using all five digits 0, 2, 4, 6, 8 once each, there are \(5! = 120\) arrangements. Those beginning with 0 are really 4-digit numbers and must be removed; there are \(4! = 24\) of them.
\[ 5! - 4! = 120 - 24 = 96 \]
Equivalently, the first digit has 4 choices (not 0) and the rest can be arranged in \(4!\) ways: \(4 \times 24 = 96\). 120 forgets to exclude a leading zero; 64 and 88 come from other miscounts.
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