# How many 4 digit numbers can be formed? [closed]

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]

1. .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.

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

## closed as too broad by Shaun, Shailesh, Leucippus, Namaste, TaroccoesbroccoAug 14 '18 at 1:30

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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: $$\begin{array}{c|c|c} \text{4-digit}&\text{1-digit}&\text{2-digit}&\text{3-digit}\\ \hline 1 & 5 \ \text{options:} \ 4,5,7,8,9 & 6 \ \text{options} \ & 5 \ \text{options}\\ 5& 4 \ \text{options:} \ 4,7,8,9 & 6 \ \text{options}& 5 \ \text{options}\\ 7& 4 \ \text{options:} \ 4,5,8,9 & 6 \ \text{options}& 5 \ \text{options}\\ 9& 4 \ \text{options:} \ 4,5,7,8 & 6 \ \text{options}& 5 \ \text{options} \end{array}\\ \text{Hence:} \ 5\cdot 6\cdot 5+4\cdot 6\cdot 5+4\cdot 6\cdot 5+4\cdot 6\cdot 5=510.$$ $$\begin{array}{c|c|c} \text{4-digit}&\text{1-digit}&\text{2-digit}&\text{3-digit}\\ \hline 0 & 5 \ \text{options:} \ 4,5,7,8,9 & 6 \ \text{options} \ & 5 \ \text{options}\\ 2& 5 \ \text{options:} \ 4,5,7,8,9 & 6 \ \text{options}& 5 \ \text{options}\\ 4& 4 \ \text{options:} \ 5,7,8,9 & 6 \ \text{options}& 5 \ \text{options}\\ 8& 4 \ \text{options:} \ 4,5,7,9 & 6 \ \text{options}& 5 \ \text{options} \end{array}\\ \text{Hence:} \ 5\cdot 6\cdot 5+5\cdot 6\cdot 5+4\cdot 6\cdot 5+4\cdot 6\cdot 5=540.$$

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 $3\cdot 6 \cdot 5 \cdot 4 + 2 \cdot 6 \cdot 5 \cdot 3$.

• Can you show that the sum of all 4-digit even and odd number that can be formed is equal to the total number of 4-digit number that can be formed using the given digits that is greater than 3000? (without repetition of digits) – MRA Aug 12 '18 at 15:25

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. $(3 \cdot 6 \cdot 5 \cdot 4)$

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.

• Careful: no repetition is allowed here. – Dan Uznanski Aug 12 '18 at 14:42
• Sorry, you made some mistakes in 2. What if you let $8$ finish the number? Then you only have 4 choices to begin. – xbh Aug 12 '18 at 14:42
• oh, you mean $8008$ counts as a repetition? I thought we're asked how many unique 4 digit numbers can be formed. – Alvin Lepik Aug 12 '18 at 14:44
• Also no repetition here. – xbh Aug 12 '18 at 14:44
• Yes. We can only use a digit once to form a 4-digit number Sir. – MRA Aug 12 '18 at 14:45