The question is:
How many ways can the numbers $0, 2, 4, 5, 7$ form a four or five digit even number greater than $4000$ if repetition is not allowed?
Can someone verify my response/method:
To form a four-digit number greater than $4000$, there must be one of the digits $0, 2, 4$ in the units position, and one of the digits $4, 5, 7$ in the thousands position, that leaves three digits for the hundreds and tens positions. Therefore there should be $3×3×2×3$ ways of arranging a four digit even number greater than $4000$.
To form a five-digit even number there must be one of the digits $0, 2, 4$ in the units position, and one of the digits $2, 4, 5, 7$ in the ten thousands position, and there would be three digits left for the thousands, hundreds and tens positions. So there would be $4×3×2×1×3$ ways of arranging a five digit even number greater than $4000$. The total arrangements would be:
$3×3×2×3$ + $4×3×2×1×3$=$126$
I am not sure how an even digit in the units position affects the overall arrangements because it may also be included as the first digit of each number.
Can someone verify my response/method?
Not correct. Note that if you start with a $4$, you can no longer end with a $4$.
To correct for this, split into cases.
For example, for the $4$-digit numbers, if you start with a $4$, then you have $2$ digits left for the last digit, and $3 \cdot 2$ for the second and third, and if you don't start with a $4$, you have $2$ options for the first, $3$ for the last, and then $3 \cdot 2$ for the second and third, for a total of $(2 + 2 \cdot 3) \cdot 3 \cdot 2$
For $5$-digit numbers, split between those that start with a $2$ or $4$, and those that start with $5$ or $7$ ... can you finish?