<s>[INST] The number of three digit numbers $abc$, which satisfy $a≤b>c$ is My approach in this question is as follows,
⇒When $a=1$
$-> 1-1-(0)  / 1-2-(0,1) / 1-3-(0,1,2) ... / 1-9-(0,1...,8)$ ⇒Total numbers = $1+2+3+...+9$
⇒When $a=2$
$-> 2-2-(0,1) / 2-3-(0,1,2) ... / 2-9-(0,1...,8)$ ⇒Total numbers = $2+3+...+9$
⇒When $a=3$
$-> 3-3-(0,1,2) ... / 3-9-(0,1...,8)$ ⇒Total numbers = $3+...+9$
So on counting, we get total numbers as,
$$(1+2+...9)+(2+3+...9)+...(8+9)+(9)$$
$$⇒1^2+2^2+3^2+...9^2$$
$$⇒\frac{(9)(9+1)(2*9+1)}{6}$$
$$⇒285$$
But the actual solution says the answer is,
$$\bigl(2*({9 \choose 1}+{9 \choose 2}+{9 \choose 3})\bigr)$$
$$⇒258$$
I think I have recounted cases, can anyone explain which cases have I recounted??
 [/INST] The given condition should be extended to
$$1\leq a\leq b>c\geq0\ .$$
Given $b\in\{1,2,\ldots,9\}$ we therefore have $b$ choices for $a$ and $b$ choices for $c$. The number of admissible three digit numbers $abc$ therefore is
$$\sum_{b=1}^9b^2={n\cdot(n+1)(2n+1)\over6}\biggr|_{n=9}={9\cdot10\cdot 19\over 6}=285\ .$$
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