2002 Math OSP, Number 18

The digits we are going to use are $1, 2, 3, 4, 5, 6, 7, 8, 9$ .

• For $a < b < c < d$ , we have

  = ${ 9 \choose 4} = 126$

• For $a < b < (c = d)$ , we have

  = ${ 9 \choose 3} = 84$

• For $a < (b = c) < d$ , we have

  = ${ 9 \choose 3} = 84$

• For $(a = b) < c < d$ , we have

  = ${ 9 \choose 3} = 84$

• For $a < (b = c = d)$ , we have

  = ${ 9 \choose 2} = 36$

• For $(a = b) < (c = d)$ , we have

  = ${ 9 \choose 2} = 36$

• For $(a = b = c) < d$ , we have

  = ${ 9 \choose 2} = 36$

• For $a = b = c = d$ , we have

  = ${ 9 \choose 1} = 9$

• The number of 4-digit numbers with $a \leq b \leq c \leq d$ are:

  $= 126 + 84 + 84 + 84 + 36 + 36 + 36 + 9$

  $= \boxed{495}$

Choose 4 numbers from the set $\left\{1,2,3,4,5,6,7,8,9,ab,bc,cd\right\}$ where [ab] denotes b=a, [bc] denotes c=b, [cd] denotes d=c.

There are $\left(\begin{array}{c}12\\ 4\end{array}\right)$ = 495 ways to do so.