The number of positive integers which can be formed by using any number of digits from 0, 1, 2, 3, 4, 5 without repetition.

Confused on selecting 2digit numbers... How and why

For those who didn't understand the above given explanation :
Now,
PERMUTATION : Order does matter in this.
COMBINATION : Order does not matter.
For example : ABC or ACB may be the same combination but different permutation.
(HERE IT'S A PERMUTATION + COMBINATION)
Formula for combination : n! / (n-r)! × r!
Formula for permutation : n! / ( n-r)!

So if I consider positive one digit no.s
Essentially they would be 5 { in counting }.
Since 0 is not a positive integer.
Now if I consider 2 digit no.s
They would be 6!/4! = 30 ways
Since here the order does not matter
i.e. no. may be 25 or 52 ,36 or 63
But the twist is that because of a zero some of the no.s may not become two digit integers
For e.g. 01,02,03 etc.
So for calculating them we will use the combination formula because 0 is constant at its place and would remain at ten's place only.
Since 0 is constant we remain with only 5 numbers to deal with so n=5 and r=1
5! / 1! × 4! =5 ways.
So now we will get the no. of ways = 30 - 5 = 25
For two digit no.s
Similarly this rule applies to all and you'll get your answer.
THANK YOU
MOKSH SAINI
CLASS 10

ok i understand this

please explain this solution in detail

for those who cant get above solution.
One digit positive numbers = 6p1-5p0=5
Two digit positive numbers = 6p2-5p1=20
Three digit positive numbers = 6p3-5p2=100
4 digit positive numbers = 6p4-5p3=300.
5 digit positive numbers = 6p5-5p4=600.
Six digit positive numbers =6p6-5p5= 600.
Total positive numbers,
= 5+25+100+300+600+600
= 1630.

How did you get this answer

how 5,25 ,300 and son is obtained?

How do you get 25

how do you get 25 . 100, 300 etc??