algebra precalculus - Find $c=atimes b$ such that $c$ has digits from either $a$ or $b$

The problem statement: Two numbers $2 \leq a, b <100$ are multiplied,
the resultant number has same digits of either $a$ or $b$
only, how many such numbers $(c)$ are there and which are they?

Example: $10 \times 10 = 100 $, on the left hand side we have $0,1$ and on the right hand side also we have $0,1$.

Since $99\times 99 =9801$, $c$ cannot be more than $4$ digits

Let $c=c_4c_3c_2c_1=c_4\times 1000+c_3\times 100+c_2\times 10+c_1$

• When both $a,b$ are one digit, there are $ {8\choose 1}\times {8\choose 1}=8\times8=64, 64-28=36$, (subtracting the repeated multiplications)




• When both $a,b$ are two digit, there are $ {90\choose 1}\times {90\choose 1}=90\times90=8100, 8100-4005=4095$, (subtracting the repeated multiplications)




• When one is one digit and second is two digits, $ {8\choose 1}+ {90 \choose 1} = 8+90=89$, (No repeated multiplications)

So, there are $36+4095+89=4220$ multiplications to be checked. What to do next?

In another approach, $c=c_4c_3c_2c_1$, since $c_4,c_3,c_2,c_1$ should either belong to $a$ or $b$

• $a$ can be formed in $4 \choose 1$ ways, $b$ can be formed in $3 \choose 2$ ways, so there are $4 \times (3\times 2!)=24$ such numbers (this case may be invalid as one digit from $c$ is left out, but what if $c$ has a digit repeated?)


• $a$ can be formed in $4 \choose 2$ ways, $b$ can be formed in $4 \choose 2$ ways, for each selection, rearrangement produces another number, so there are $6\times 2!=12$ such numbers.

How to find these $36$ numbers ?