How many integers could be in such a way that any digits is not bigger than the left digits?

Question

How many 4-digits integers could be in such a way that any digits is not bigger than it's left digits?

I Try it with simulation, i get 714. anyone could describe a formula for me?

My try:

Answer 1

Assuming you mean the four digit number to have the digits weakly decreasing (each less than or equal to the digit to the left), I get $715$, so think you missed one. If I had to guess it is $0000$, but maybe you don't allow that. One way to think of it is to select four digits allowing repetition. Each selection generates a single solution-just sort the digits selected in descending order. One way to get this is stars and bars. Add 3 to the thousands digit, 2 to the hundreds, 1 to the tens, and 0 to the ones. Now we don't allow any ties. Choose 4 numbers from the range 0 through 12 without repetition, which you can do in ${13 \choose 4}=715$ ways. Sort them, subtract the numbers added, and you have a four digit number with the digits weakly decreasing.

Another approach is to enumerate the distributions of duplicate digits. No duplicates gives ${10 \choose 4}=210$ One pair gives $10{9 \choose 2}=360$ A triplet gives $90$. Four of a kind gives $10$. Two pair gives ${10 \choose 2}=45$ and $210+360+90+10+45=715$