# Find. total number of $3$ digit of the form $xyz$, where $x<y>z$

Finding total number of $$3$$ digit integer of the form $$xyz$$ (all digit distinct) by using the number $$\{0,1,2,3,4,5,6,7,8,9\}$$ which are is in the form

$$(1)\ x

$$(2)\ x\leq y

$$(3)\ xz$$

$$(1)$$ For first case

Selecting $$3$$ distinct numbers out of total $$10$$ numbers

$$\displaystyle =\binom{10}{3}$$

$$(2)$$ For second case

We have either $$x or $$x=y

So total possibility $$\displaystyle =\binom{10}{3}+\binom{10}{2}=\binom{11}{3}$$

Please find where I have solve above two part correctly or not and also please have a look on third part

• Can $x$ be $0?$ Mar 3 at 3:20
• Don't write things like "$x\lt y\gt z$." The proper way to write it is either "$x\lt y$ and $z\lt y$", or "$x,z\lt y$". Mar 3 at 3:21
• The condition "all digits distinct" is contradicted by $x\leq y$ Mar 3 at 3:21
• Your answers are correct as long as "3-digit integer" includes ones with a leading zero (e.g. 027); for the 3rd case, use a similar approach as for Part 2: either x = z or x $\neq$ z.
– A.J.
Mar 3 at 4:01
• For question $3$, step1, how many options to select $3$ distinct digits, and step2, when you have $3$ distinct digits, how many options to organize them, with the constraints $x<y$ and $z<y$ ? Mar 3 at 8:11

1 - The selection must be done from the numbers $$1 - 9$$ since no number can have a $$0$$ in this case

2 - Again , the selection must be done from numbers $$1-9$$, moreover two numbers that are equal to each other along with a bigger number need to be selected to satisfy the equality part. So the options for $$x$$ and $$y$$ are from $$1-8$$ and in each selection of $$x$$, $$z$$ has $$9-x$$ options. Hence the total options are $$8 + 7+ 6+5+4+3+2+1$$ i.e $$36.$$

3- For any value of $$y$$, $$z$$ can possess $$y$$ choices (excluding $$y$$ and including $$0$$) and x can possess $$y-1$$ choice(excluding $$y$$ and $$0$$). Thereby total options for the formation of XYZ is $$\Sigma_{i=1}^9 I (i-1)$$ which is $$240$$.

I would start with $$y$$ and go on from there:

1. $$y = 0$$ : not possible
2. $$y = 1$$ : result does not have three digits
3. $$y = 2$$ : 1 possibility for $$x$$ and 2 for $$z$$
4. $$y = 3$$ : 2 possiblities for $$x$$ and 3 for $$z$$

...

1. $$y = 9$$ : 8 possibilities for $$x$$ and 9 for $$z$$

So, you get:

$$1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5 + 5 \cdot 6 + 6 \cdot 7 + 7 \cdot 8 + 8 \cdot 9 = 240$$