# How many numbers between 4,000 and 7,000 can be chosen using the digits [0, 8]?

I have a homework problem in combinatorics, and I am struggling to solve it because I didn't understand our lesson well.

How many numbers between 4,000 and 7,000 can be chosen using the digits 0, 1, 2, 3, 4, 5, 6, 7, and 8 if each digit must not be repeated in any numbers?

PS: I don't have the resources to solve this. I just don't understand what our professor taught us, but we have already made a make-up class to solve this conflict.

• The idea is to determine how many choices you have for each digit given the constraints. The number $7000$ does not satisfy the given conditions since $0$ is repeated. Thus, you need only choose numbers between $4000$ and $6999$ that satisfy the given conditions. If you make an attempt, you are more likely to have your question answered. – N. F. Taussig Aug 12 '15 at 11:25

HINT...you have a choice of 4, 5 or 6 for the first place (from left to right).

You then have a choice of 8 digits for the second place (including 0 but excluding your first choice).

Then you have a choice of 7 digits for the third place..... can you finish this?

• Thank you for answering but, I have a question , Is it related to Combination and Permutation? Because our lesson was all about it. Does a formula applies on this question? I just confused, im very sorry :( ill is this the formula? (4x5x6)x(8!)x(7!) ? – jAvA tHe Gamer Aug 12 '15 at 11:42
• More precisely, It is related to arrangements, also known as $k$-lists. – Bernard Aug 12 '15 at 12:26
• SO i Got it, Please tell me if my answer is correct: So in first digit (in left) i have a 3 choices (4,5,6) So my answer should be 3x8x7x6 = 1008? – jAvA tHe Gamer Aug 12 '15 at 12:37
• @jAvAtHeGamer. yes that's right, and we're talking permutations not combinations, since order matters – David Quinn Aug 12 '15 at 13:39

There are some correct answers but they don't explain well enough for my taste.

First, to think about these types of problems, it helps to visualize what is going on. In this case, I would use four spaces to visualize a four digit number. $$_ _ _ _$$ Now we want to figure out how many ways we can fill in each blank, being very careful to count each possibility once and only once. First(and I am not going into full detail here) we usually want to start with the MOST restricted choice. In this case, that is the first digit. Since 7000 isn't allowed due to repetition, we can have either 4,5 or 6 as the first digit. Thus there are 3 possibilities. Now if we had a number in the first place(for the sake of visualization say 4) $$4 _ _ _$$ Now how many choices do we have for the next digit? There are 8 allowable digits we haven't used yet, so eight ways we can fill in the second digit(did this depend on our choice of using 4 as the first digit?)

What about after that? The third digit now has seven possibilities and the fourth is left with six. Thus we can count this as $$3*8*7*6=1008$$ It should also be noted that this is counted by $$3*P(8,3)$$ since the last three digits can be thought of as lining up three objects from 8 since order matters, repetition is not allowed and there are no other restraints on the choices(the three things to check for before using a permutation formula)

• well, very appreciate dude. I clearly understand the concept because u use visualize. THANK YOU – jAvA tHe Gamer Aug 12 '15 at 12:26

You first have to choose the first digit ($\neq 0,9$), then choose any $3$ distinct digits among the remaining ones, taking into account their order. This is called the number of $3$*-arrangements* on the set of remaining digits.

Let's compute all this:

General case: the first digit can be any digit, but $0$ and $9$. Hence there are $8$ possible choices. For the other digits you need to choose a list of $3$ distinct digits among the $8$ remaining digits. There are $$A_8^3=\frac{8!}{(8-3)!}=8\cdot7\cdot6,\enspace\text{whence}\quad 8^2\cdot7\cdot 6\enspace\text{possibilities}.$$

Favourable case: the first digit must be one of $4,5,6$, i.e. there are $3$ possible choices. The other digits are as above, whence $$3\cdot8\cdot7\cdot6=1008 \enspace\text{favourable cases.}$$

If we choose as sample space the set of quadruples of digits between $0$ and $8$, the first digit being different from $0$, the probability of having a number with distinct digits in $\{\,0,\dots,8\,\}$, between $4000$ and $7000$, is equal to $\dfrac38$.

Choosing as a sample space the set of all $4$-digit numbers, the answer is different for the general case: we have $9000$ possibiliites in all, so the probability is: $$\frac{3\cdot8\cdot7\cdot6}{3^2\cdot 2^3 \cdot5^3}=\frac{14}{125}.$$

• The constraints of the problem don't allow for using the digit 9 so you are over counting here. – Sean English Aug 12 '15 at 12:09
• Oh! Forgot that. I'll correct my answer. – Bernard Aug 12 '15 at 12:12
• The question only asks the count, not the probability. You might consider deleting that portion. – true blue anil Aug 12 '15 at 12:21
• Yes but I thought it might be of interest, as it shows what is important is only the choice of the first digit. – Bernard Aug 12 '15 at 12:27
• If you are going to include the probability section, I would recommend making it very clear as to what your sample space is since there are many choices that we could look at, and the most obvious one(in my mind) isn't the one you are using(this being all the numbers from 4000 to 7000.) – Sean English Aug 12 '15 at 12:31

You got $3 (4,5,6)$ choices for the first place.

8 choices for the second place, 0-8 and excluding the one number you already chose for the first place.

7 for third one and 6 for the last.

Overall, using this you can make $3*8*7*6=1008$ such numbers.

The first digit can be 4,5,6 footie that we exclude 7 because 7000 repeats 0, so there are 3 possible choices for the first digit. Then you have 8 choices (0,1,2,3,4,5,6,7,8 excluding the first digit) for the next digit, and each following digit has one less. So you get 3*8*7*6=1008. So there are 1008 possible numbers.