Question

How many positive integer numbers not more than 4300 can be formed with the digits 0, 1, 2, 3, 4 if repetitions are allowed?

Answer 1

Answer:

575

Step-by-step explanation:

First the number should be equal to or less than 4 digited. I'll write down all the cases involved in the question.  

Note: repetition is allowed!!

1) Three digit numbers:

The first digit can't be zero. So it has 4 chances, and the remaining two has 5 chances each. So number of 3-digit numbers are 5*5*4=100.

2) Two digit numbers:

The first place has 4 chances and the second one has 5. So 20  two digit numbers can be formed.

3) One digit numbers:

It is obvious that 5 one digit numbers can be formed.  

4) four digit numbers:

4-a) Fix 3 in thousands place:

we have 5 chances for the remaining three places. so we have 125 chances.

4-b) Fix 2 in thousands place:

The answer is the same as in the previous one!!!

4-c) Fix 1 in thousands place:

The answer is the same as in the previous one!!!

4-d) Fix 4 in thousands place:

Since the number should be less than 4300, we can't have 4 and 3 in hundreds place.

So we have 3*5*5 chances in this case.

Total chances so far are 100+20+5+125+125+125+75.

As per the given conditions, total numbers are 575!!!

Hope this helps!!