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How many three digit numbers of the form $xyz$ are there such that $x<y>z$?

So, it's obvious from the question that $x \ne 0$, so even $y\ne0$.

My approach: I counted.

$x$ $y$ $z$
$1$ $2$ $0,1$
$1,2$ $3$ $0,1,2$
$1,2,3$ $4$ $0,1,2,3$
$1,2,3,4$ $5$ $0,1,2,3,4$
$1,2,3,4,5$ $6$ $0,1,2,3,4,5$
$1,2,3,4,5,6$ $7$ $0,1,2,3,4,5,6$
$1,2,3,4,5,6,7$ $8$ $0,1,2,3,4,5,6,7$
$1,2,3,4,5,6,7,8$ $9$ $0,1,2,3,4,5,6,7,8$

After I counted all cases, I got:

$$80+70+60+50+..+10$$ which gives $360$. But the answer given is $240$. Can someone confirm if my method is correct? Is there a shorter method to do this?

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  • $\begingroup$ Where are you getting the numbers in your sum? For example, the only qualifying numbers I see with $y=2$ are $121$ and $120$. I think that's probably your mistake. $\endgroup$ May 12, 2021 at 9:09

3 Answers 3

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You have to count all possible combinations of numbers from both first and third column in your table. For example, when $y=2$, you have only one possible value of $x$, and two possible values of $z$. Thus, the number of possible combinations is $1*2=2$. Proceeding the same way with the rest of the rows, we obtain:

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

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Using your method, the number is

$$\sum_{i=1}^n i(i+1)=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)}{6}(2n+4)=\frac{8(9)(10)}{3}=240$$

where $n=8$.

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I think you could do it in an easier way. Take two cases:

  1. $x$,$y$, and $z$ are distinct. Choose these from the set $A=\{1,2,3,...,9\}$. You have $\binom {9}{3}$ choices. Once chosen, each combination gives $2$ numbers. So, you have $2 \binom {9}{3}$ numbers. Now, take the case where $z=0$ and choose $x,y$ from the set $A$. I leave the calculation to you, it is quite elementary and in the same manner as I showed.
  2. Where $x=z$. Here, choose $2$ numbers from $A$, and for every choice, there is only one $3$-digit number that can be formed from these digits satisfying the given condition(once you've chosen two numbers, the greater becomes $y$ and the smaller one is $x=z$. Arrangement can only be done in one way.

Add the two cases, you get your total.

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