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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

$(1)\ x<y<z$

$(2)\ x\leq y<z$

$(3)\ x<y>z$

$(1)$ For first case

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

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

$(2)$ For second case

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

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

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    $\begingroup$ Can $x$ be $0?$ $\endgroup$
    – Thomas Andrews
    Mar 3 at 3:20
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    $\begingroup$ 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 "$x,z\lt y$". $\endgroup$
    – Arturo Magidin
    Mar 3 at 3:21
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    $\begingroup$ The condition "all digits distinct" is contradicted by $x\leq y$ $\endgroup$
    – Thomas Andrews
    Mar 3 at 3:21
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    $\begingroup$ 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. $\endgroup$
    – A.J.
    Mar 3 at 4:01
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    $\begingroup$ 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 $z<y$ ? $\endgroup$
    – Lourrran
    Mar 3 at 8:11

2 Answers 2

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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 $240$.

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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:

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

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