# 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?

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?

• Split the cases up as 1/ Leading digit is 4, 5, or 6, 2/ Leading digits are 34, 35, or 36, 3/ Leading digits are 32. What would you get for those? Commented Mar 27, 2023 at 15:43
• @Calvin: and do these as fractions of $6!$, so $\frac36+\frac45 \times \frac16$ Commented Mar 27, 2023 at 15:47

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 $$720-264=456$$

• Each starting digit gives $$5$$ possible two digit starts$$\,\rightarrow 30$$ total

• Five each starting with $$1$$ or $$2$$ are "bad" as is one starting with $$32$$, leaving $$30-11=19$$ "good" two digit starts

• Thus there are $$19\times4! = 456$$ "good" numbers

• but what about 41xxxx, 51xxxx and 61xxxx? Commented Mar 28, 2023 at 5:05
• @somebody: Good catch, corrected ! Commented Mar 28, 2023 at 6:23
• What's a "five each"? @trueblueanil
– user1165756
Commented Apr 1, 2023 at 0:16
• @Lasuiqw: Edited for greater clarity, "Five each" two digit starts Commented Apr 1, 2023 at 5:58