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


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 $210+360+90+10+45=715$
