Given just a $4$-distinct-digit number. What are the odds that randomly chosen number has digit $1$ and digit $2$ ?
My own observation
With $4$ digits $1, 2, 3, 4$ ; $1$ in $1$
With $5$ digits $1, 2, 3, 4, 5$ ; $3$ in $5$
With $6$ digits $1, 2, 3, 4, 5, 6$ ; $6$ in $15$
I guess..
With $9$ digits $1, 2, 3, \cdots, 9$ ; $\binom{7}{2}$ in $\binom{9}{2}$
With $10$ digits $1, 2, 3, \cdots, 9, 0$ ; how ?
I need to more hints, thank you very much. Also, I want to see a cleverer way(s)..
There are $9*9*8*7$ possible $4$ (distinct) digit numbers ($9$ choices for 1st spot, etc)
Count number containing both $1$ and $2$:
A) Pick a spot for digit $1$ ($4$ choices)
B) Pick a spot for digit $2$ ($3$ choices)
Now we split up based on whether first spot is used yet. Exactly $6$ of the $12$ choices so far have a digit in the first spot.
C1) If first spot taken, then $8*7$ choices for the two available spots in order. This option occurs in $6$ of the $12$ choices in (A) and (B).
C2) If first spot open, then $7$ choices for first spot, $7$ choices for last open spot. This option occurs in $6$ of the $12$ choices in (A) and (B)
So: $(6*56 + 6*49) / (9*9*8*7)$