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Hello.
00:01
So here, our first question is, we want to know how many four decimal digit numbers are there between 1 ,000 and 9 ,9009 that do not contain the same digit twice.
00:16
So here, if we're not going to contain the same digit twice, or then we have first, we have nine choices, one to nine, and then we have again, nine choices.
00:29
But then we have, if we're not going to contain the same choice, our last two choices are going to be eight choices and then seven choices.
00:36
So therefore, we're going to have nine times nine times eight times seven.
00:40
So it's going to be a total of 4 ,536 numbers.
00:44
And then for part b, we want to know if we end with an odd digit.
00:49
So if we end with an odd digit, so our first digit in our number, we have nine choices, either one to nine.
00:58
But then we're going to have 10 choices for the second number, right? because this is going to be through 9 ,999.
01:04
But the second digit could be any digit at all.
01:07
So we're going to have 10 choices, 0, 1, 2, 3, 4, 6, 7, 8, or 9.
01:12
So we'd have 9 times 10.
01:14
And likewise, for the third digit, we'd have 10 choices.
01:17
But then if we end with an odd digit, then our last digit only has five choices because there's five odd numbers, either 1, 3, 5, 7, or 9.
01:26
So therefore we have nine times 10 times 10 times five...