Permutation and combinatorics problem

Question

How many numbers between 10 and 1000 can be formed using digits 3,4,5,7?

Should we first find number of 2 digit numbers and then find number of three digit numbers and add them

Answer 1

All $3$ and $2$ digit numbers that can be formed from these digits lie between $10$ and $1000$.

Number of $3$ digit numbers:

$$\underbrace{\binom{4}{3}}_{\text{select any 3 digits}}\times\underbrace{3!}_{\text{Permutations}} = 4\times6=24$$

Find number of $2$ digit numbers similarly, and add.

Answer 2

Yes, your approach

Should we first find number of 2 digit numbers and then find number of three digit numbers and add them

is correct. You have that

1. For a 2-digit number. You have 4 choices for the first digit and (since repetetition is allowed) 4 choices for the second digit. That gives you $$4\times 4=16$$ different 2-digit numbers.
2. For a 3-digit number. Similarly, you have 4 choices for the first digit, 4 choices for the second digit and4 choices for the third digit. That gives you $$4\times 4 \times 4 =4^3=64$$ different 3-digit numbers.

Adding the two results you have that there are $$16+64=80$$ different numbers between 10 and 1000 with the digits 3,4,5 and 7.

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Note: You should have been more carefull if the digit $0$ was among the $4$ possible digits, since then $0$ would not have a possible choice for the first digit, but this was not the case here.