Question

Permutation and Combinations Question:
How many even four digit numerals have no repeated digits?

Answer 1

There are \(10\) digits: \(0\), \(1\), \(2\), \(3\), \(4\), \(5\), \(6\), \(7\), \(8\), \(9\).

Therefore, a four-digit, even numeral with no repeated digits has \(10\cdot9\cdot8\cdot5\) possibilities, whenever the first three digits are all odd, \(9\cdot8\cdot4\) whenever one of the first three digits is even, \(8\cdot3\) whenever two of the first three digits is even, and \(2\) whenever the first three digits are all even. As a result, you have \(3914\) possibilities total.

Answer 2

I appreciate your reply.
Though, I still don't get it. I have an answer key and it is 2296. I don't know they arrived to that.

Answer 3

how*

Answer 4

you have 4 places and 5 digits to fill in
the first place cant take 0. so it has 4 choices
the next place cannot take the one in the first place, so it can take 4 digits
similarly proceed