How many even numbers can you form greater than $5000$ from the digits $1$ to $7$ inclusive, if no repetition is allowed?
My Attempted Solution
$n = 3 \times 6 \times 5 \times 3 = 270$
My reasoning, for the first digit, only $5, 6, 7$ can be chosen, thus leading to $3$ possible choice, and since no repitition is allowed it follows that there can only be $6$ subsequent choices for the second digits and $5$ for the third. Finally only $2, 4, 6$ can be chosen for the fourth digits, leading to $3$ possible choice for the fourth digit.
Is my solution correct? If not, where is the flaw in my logic?
Best Answer
I believe you have a flaw in your logic. There's no way of knowing how many even digits you have for the last digit.
I would probably break this problem into two cases. The first is that the first digit is a $6$. I'd then consider what the last digit could be since it's the only other digit that has restrictions. Then the remaining digits can be chosen from whatever's left over.
Upon seeing Chas Brown's comment bringing up that the problem has apparently not forbidden numbers with five or more digits, I decided I should address those cases as well. The simplest method adds another three cases: one for five-digit, six-digit, and seven-digit numbers. These three cases should be simpler since the thousands digit no longer matters. There are three ways to choose the final digit while each remaining digit can be anything that hasn't already been chosen.