How many 4 digit numbers can be formed?

Question

Given digits: 0, 1, 2, 4, 5, 7, 8 and 9

1.How many 4-digit numbers can be formed greater than 3000 without repetition? [Here we mean no repetitive digits]

My answer is 5*7*6*5

2. .How many 4-digit even numbers can be formed greater than 3000 without repetition?

My answer is sum of 2*6*5*4 and 3*6*5*4.

3. Why should we not just divide the answer in no.1 by two to get the answer in no.2?

Answer 1

1) correct. 2) it must be $2\cdot 6\cdot 5\cdot \color{red}{3}+3\cdot 5\cdot 6\cdot 4$. Interpretation: two cases: $$\begin{array}{c|c|c} \text{1-digit}&\text{4-digit}&\text{2-digit}&\text{3-digit}\\ \hline \text{even} \ (2 \ \text{options:} \ 4,8) & \text{even} \ (3 \ \text{options:} \ 0,2,4 \ \text{or} \ 8)& 6 \ \text{options}& 5 \ \text{options}\\ \text{odd} \ (3 \ \text{options:} \ 5,7,9)& \text{even} \ (4 \ \text{options:} \ 0,2,4, 8)& 6 \ \text{options}& 5 \ \text{options}\\ \end{array}$$ 3) because there are $510$ odd and $540$ even numbers: $$\begin{array}{c|c|c} \text{4-digit}&\text{1-digit}&\text{2-digit}&\text{3-digit}\\ \hline 1 & 5 \ \text{options:} \ 4,5,7,8,9 & 6 \ \text{options} \ & 5 \ \text{options}\\ 5& 4 \ \text{options:} \ 4,7,8,9 & 6 \ \text{options}& 5 \ \text{options}\\ 7& 4 \ \text{options:} \ 4,5,8,9 & 6 \ \text{options}& 5 \ \text{options}\\ 9& 4 \ \text{options:} \ 4,5,7,8 & 6 \ \text{options}& 5 \ \text{options} \end{array}\\ \text{Hence:} \ 5\cdot 6\cdot 5+4\cdot 6\cdot 5+4\cdot 6\cdot 5+4\cdot 6\cdot 5=510.$$ $$\begin{array}{c|c|c} \text{4-digit}&\text{1-digit}&\text{2-digit}&\text{3-digit}\\ \hline 0 & 5 \ \text{options:} \ 4,5,7,8,9 & 6 \ \text{options} \ & 5 \ \text{options}\\ 2& 5 \ \text{options:} \ 4,5,7,8,9 & 6 \ \text{options}& 5 \ \text{options}\\ 4& 4 \ \text{options:} \ 5,7,8,9 & 6 \ \text{options}& 5 \ \text{options}\\ 8& 4 \ \text{options:} \ 4,5,7,9 & 6 \ \text{options}& 5 \ \text{options} \end{array}\\ \text{Hence:} \ 5\cdot 6\cdot 5+5\cdot 6\cdot 5+4\cdot 6\cdot 5+4\cdot 6\cdot 5=540.$$

Answer 2

1. The first digit has to be at least $4$, this gives us five options. The rest simply can't repeat so by product rule there are $7\cdot 6\cdot 5$ options to complete the number. In total $7\cdot 6\cdot 5^2$ choices.
2. First digit is still at least $4$. There are two cases to consider.

If the first digit is odd (three choices), then there are four ways to pick the last digit. To avoid repetition in between there are in total $3\cdot 4\cdot 6\cdot 5$ choices.

If the first digit is even (two choices), then there are three ways to pick the last digit. Avoiding repetition in between there are in total $2\cdot 3\cdot 6\cdot 5$ choices.

There is obviously no overlap between numbers starting with even/odd digit so one sums up the number of choices for the total.

Answer 3

For your third question, to do that, you are assuming that there are equal number of odd and even number of $4$ digits number formed by those particular digits that are greater than $3000$. You have to justify if this assumption is valid.

For your second question. We need to pay attention to the first and the last digit.

If the first digit is even, there are $2$ choices. After using an even digit at the first place, you have two choices at the last position. $(2 \cdot 6 \cdot 5 \cdot 3)$.

If the first digit is odd, there are $3$ choices. There are $3$ options at the last digit. $(3 \cdot 6 \cdot 5 \cdot 4)$

Answer 4

The reason it's not merely half is because, by picking the thousands digit first, we restrict the ones digit in different ways: if we select an odd thousands digit (of which there are three), there are four legal choices for the ones digit, but if we select an even thousands digit (of which there are two), there are only three.

Which is to say, the number of numbers that can be formed is in fact $3\cdot 6 \cdot 5 \cdot 4 + 2 \cdot 6 \cdot 5 \cdot 3$.