How many 6-digit integers greater than 321,000 can be formed such that each of the digits 1, 2, 3, 4, 5, and 6 is used once in each 6-digit integer?
Keep original(fix) the hundred thousands, ten thousands, and thousands place (all permutations of the hundreds,tens, and ones is $3!.$
Keep original(fix) hundred thousands, ten thousands, and the thousands place to two (all permutations of the hundreds,tens, and ones is $3!.$
Keep original(fix) the hundred thousands, ten thousands, but the thousands place to four (all permutations of the hundreds,tens, and ones is $3!.$
Keep original(fix) the hundred thousands, ten thousands, but the thousands place to five (all permutations of the hundreds,tens, and ones is $3!.$
Keep original(fix) the hundred thousands, ten thousands, but the thousands place to six (all permutations of the hundreds,tens, and ones is $3!.$
So far we have $ 5 \cdot 3!$
Keep original(fix) the hundred thousands, but ten thousands is $4$, the thousands (all permutations of the hundreds,tens, and ones is $3!.$
Keep original(fix) the hundred thousands, but ten thousands is $5$, the thousands (all permutations of the hundreds,tens, and ones is $3!.$ ..
Keep original(fix) the hundred thousands, but ten thousands is $6$, the thousands (all permutations of the hundreds,tens, and ones is $3!.$ ..
Keep original(fix) the hundred thousands, but ten thousands is $4$, the thousands (all permutations of the hundreds,tens, and ones is $3!.$
Keep original(fix) the hundred thousands, but ten thousands is $5$, the thousands (all permutations of the hundreds,tens, and ones is $3!.$ ..
Set the hundred thousands to four, fix the ten thousands, the thousands (all permutations of the hundreds,tens, and ones is $3!.$ ..
Set the hundred thousands to five, fix the ten thousands, the thousands (all permutations of the hundreds,tens, and ones is $3!.$ ..
Set the hundred thousands to six, fix the ten thousands, the thousands (all permutations of the hundreds, tens, and ones is $3!.$ ..
This is a bad system.
There are cases I'm not enumerating, and I don't have a system to enumerate them that allows me to multi-task and count multiple things at the same time. How to continue to capture all cases?
Best Answer
You are overcomplicating things. You need to permutate the numbers $1, 2, 3, 4, 5, 6$ to form a 6 digits number. There are two ways to count how many are greater than $321\ 000$ : directly or by counting those smaller.
Directly
Your idea is good, but you have to many cases and it is hard to keep track of what you are doing.
Hundred thousand $4, 5$ of $6$, then the rest of the number doesn't matter, it will be greater. $$3\times 5!$$ Hundred thousand $3$, ten thousand $4, 5$ or $6$, then the rest of the number doesn't matter, it will be greater. $$1\times3\times 4!$$ Hundred thousand $3$, ten thousand $2$, then the rest of the number doesn't matter, it will be greater. (The smallest such numbers is $321\ 456$) $$1\times1\times 4!$$ Total $$3\times 5!+1\times3\times 4!+1\times1\times 4! = 456$$
Counting those smaller
There is $6! = 720$ 6 digits numbers made with $1, 2, 3, 4, 5, 6$. How many of those are smaller than $321\ 000$? Hundred thousand $1$ or $2$, then the rest of the number doesn't matter, it will be smaller. $$2\times 5!$$ Hundred thousand $3$, then the ten thousand need to be $1$, the rest of the number doesn't matter.. $$1\times1\times 4!$$ Total smaller numbers $$2\times 5! + 1\times1\times 4!=264$$ Answer $$720-264=456$$