How many $5$ digit numbers are there such that digits are in ascending order ?
I think it must be $5 \times 6^4$ ,
But I'm not sure. I would love to see an algorithm to solve this kind of problems.
[Math] 5 digit numbers in ascending order
combinationscombinatorics
Best Answer
There are two answers here, depending on the criteria you hold.
If you require that the digits are strictly ascending, i.e., repeated digits are not allowed, this question becomes rather simple, as for each combination of digits, there is only "valid" ordering of numbers. Hence, we simply take the combination of 5 digits from the numbers 1,2,3,4,5,6,7,8,9 (no 0 because a number can't start with 0). This gives us $${9\choose 5} = 126$$
If you allow repeated digits, then it's a bit more tricky. If the digits are $a,b,c,d,e$, we define $u=a-1, v=b-a,w=c-b, x=d-c,y=e-d, z=9-e$. This is useful because $$u+v+w+x+y+z=8$$ and each of these numbers must be $\geq0$. So, we use stars and bars to get ${13\choose 5} = 1287$.
P.S. You could also use stars and bars for the strictly ascending case! You should try it out to see whether you can replicate the same solution I've derived.