arrangement of digits

Question

With how many ways can we arrange the digits $1,2, \dots, 9$,so that $1$ precedes $2$ and $2$ precedes $3$? Also,with how many ways can we arrange these digits,so that between $1$ and $2$ there are three digits?

Answer 1

Take a random permutation of our $9$ numbers, with all permutations equally likely.

The probability that $1$, $2$, and $3$ will be in that (relative) order is $\frac{1}{3!}$. So if $N$ is the number of permutations of the $9$ numbers in which $1,2,3$ are in that relative order, we have $$\frac{N}{9!}=\frac{1}{3!}.$$ Now we know $N$.

The second problem is ambiguous, since it is not clear whether $1,2,3$ are to be in that relative order. We assume first that the problem is a separate one.

Then the leftmost of $1$ or $2$ can be in any one of $5$ places. Whether that place is occupied by $1$ or $2$ can be chosen in $2$ ways, and the rest of the slots can be filled in $7!$ ways, for a total of $(5)(2)(7!)$.

If we assume that the condition about the ordering of $1,2,3$ must be respected, draw a little diagram, putting down $9$ slots. If $1$ is in slot $1$, then $3$ has $4$ places it can go. If $1$ is in slot $2$, then $3$ has $3$ places to go, and so on. That fives $4+3+2+1$ ways to place $1,2,3$. For each of these ways, the remaining slots can be filled in $6!$ ways.

Answer 2

For the first part I'm guessing you meant "2 precedes 3" and not "2 proceeds 3" as those are very different.

I'll show some patterns. The $1$ and $2$ will be shown, then the possible positions of $3$ will be $c$. Any other number will be $x$. These will then show how many ways $3$ can be placed after an equals sign. $$12ccccccc = 7$$ $$1x2cccccc = 6$$ $$1xx2ccccc = 5$$

and so on.

$$x12cccccc = 6$$ $$x1x2ccccc = 5$$ $$x1xx2cccc = 4$$

You get the idea. This goes until

$$xxxxx12cc = 2$$ $$xxxxx1x2c = 1$$ $$xxxxxx123 = 1$$

So we have $$\left(7+6+5+\ldots+1\right)+\left(6+5+4+\ldots+1\right)+\ldots+\left(2+1\right)+\left(1\right)$$

This is the same as $$7\cdot1+6\cdot2+5\cdot3+4\cdot4+3\cdot5+2\cdot6+1\cdot7$$

Which is the same as $$\sum_{i=1}^{7}(8-i)i = 84$$

Another way to think of it is $$\frac{7\cdot8}{2}+\frac{6\cdot7}{2}+\frac{5\cdot6}{2}+\ldots+\frac{2\cdot3}{2}+\frac{1\cdot2}{2} = \sum_{i=1}^{7}\frac{n(n+1)}{2} = 84$$

So there are $84$ ways to arrange the $1$, $2$, and $3$. So now you need to arrange the others around them. $6$ digits remain, so there are $6!$ ways to do it. $$84\cdot6!=60480$$

Note that this is WAY more complicated than some of the other solutions posted. I just wanted to show an alternate way of doing things.

For the second part, you have three digits between $1$ and $2$, giving you $1xxx2$. This arrangement is then shifted along the line of numbers. With a length of 5, that gives $$9-5+1=5$$ places this arrangement can go, the $9$ coming from the total length of the digit string. However that's for $1xxx2$. We can have $2xxx1$ and that would still count as 3 digits between $1$ and $2$. How many ways can we arrange those? $2\choose{1}$, which is $2$.

$$5\cdot\binom{2}{1}=10$$

For each of these, there are $7$ other digits to be placed, so that gives you $7!$ $$10\cdot7! = 50400$$

Answer 3

The first

Choose three places out of $9$ to put $1,2,3$ at. Then scramble the other six digits and put them into the remaining places. This can be done in $$ \binom{9}{3}\cdot 6!=60480\text{ ways} $$

The second

The digit $1$ has to go in one of the first five places otherwise there is no room for three digits and a $2$. For each such choice scramble the other seven digits and fill them in. This makes $$ 5\cdot 7!=25200\text{ ways} $$

Edit

Having read André Nicolas answer I just realized that I misread the second question. Since $1$ and $2$ may be interchanged the answer will we twice as big. so $50400$ ways it is.

Answer 4

For the first one, there are 9!/3!=60480 ways to arrange the other 6 numbers. The numbers 1,2 and 3 can now only go in one way for each of these permutations. So the total is 60480.