2 years, 2 months ago
How many numbers between 2000 and 5000 can be made from the digits 1,2,4,5,7 and 8 if each digit can be used only once?
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M$3 Answers
This problem is really simple.
You are going to be making a 4 digit number from 6 digits that can't be reused.
There are 2 possibilities for the first digit, 2 or 4.
There are 5 possibilities for the second digit (6 - 1 = 5 because you started with six and already used one)
There are 4 possibilities for the third digit. (6 - 2 = 4)
There are 3 possibilities for the fourth digit. (6 - 3 = 3)
So now just multiply the number of possibilities together:
2*5*4*3 = 120
The other answers are completely correct as far as I can tell, just overly complicated.
You are going to be making a 4 digit number from 6 digits that can't be reused.
There are 2 possibilities for the first digit, 2 or 4.
There are 5 possibilities for the second digit (6 - 1 = 5 because you started with six and already used one)
There are 4 possibilities for the third digit. (6 - 2 = 4)
There are 3 possibilities for the fourth digit. (6 - 3 = 3)
So now just multiply the number of possibilities together:
2*5*4*3 = 120
The other answers are completely correct as far as I can tell, just overly complicated.
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M$
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Here is a simple solution that requires solving 2 smaller sub-problems and then combining them to come up with a final solution:
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(P A R T . 1)
How many numbers are between 2000 and 3000 (using 1, 2, 4, 5, 7, and 8 only once)? This requires you to solve this puzzle:
2 ? ? ?, where each question mark represents 1 of 6 possible digits
BUT since we can use each digit once, we cannot use 2. So there are only 5 digits left to fill in the question marks (1, 4, 5, 7, 8). We will now simplify our possible options by using an * to denote the each of the five choices (*, *, *, *, *).
So to fill the first question mark we use one of the 5 choices:
2 * ? ?
Now we have 4 choices left (*, *, *, *).
Fill in the next question mark with one of the choices.
2 * * ?
Now we have 3 choices left (*, *, *).
Fill in the last question mark.
2 * * *
As you can see to we had 5 choices for the first digit, 4 choices for the second digit, and 3 choices for the last digit. Simply use the PRODUCT RULE to find your answer.
5 X 4 X 3 = 60
There are 60 numbers between 2000 and 3000 using 1, 2, 4, 5, 7, and 8 only once.
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(P A R T . 2)
3 is not one of the possible digits so we now ask ourselves, how many numbers are there between 4000 and 5000 (using 1, 2, 4, 5, 7, and 8 only once)? This requires you to solve this puzzle:
4 ? ? ?, where each question mark represents 1 of 6 possible digits
BUT this time lets use LOGIC to determine this answer.
4 is already used so that leaves only five digits (1, 2, 5, 7, and 8). Simplifying our process makes the 5 choices (*, *, *, *, *).
Hmmm...
Since there are only 5 choices again to fill in 3 question marks our answer will be the same as the answer in part 1. But so you get some practice go ahead and use the PRODUCT RULE again:
5 X 4 X 3 = 60
There are 60 numbers between 4000 and 5000 using 1, 2, 4, 5, 7, and 8 only once.
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(P A R T . 3)
Combining the answers from part 1 and part 2 yields:
60 + 60 = 120
Thus, there are a total of 120 numbers between 2000 and 5000 using 1, 2, 4, 5, 7, and 8 only once.
Here is a simple solution that requires solving 2 smaller sub-problems and then combining them to come up with a final solution:
--------------------------------------------------------------------------------------------------------------------------------------------
(P A R T . 1)
How many numbers are between 2000 and 3000 (using 1, 2, 4, 5, 7, and 8 only once)? This requires you to solve this puzzle:
2 ? ? ?, where each question mark represents 1 of 6 possible digits
BUT since we can use each digit once, we cannot use 2. So there are only 5 digits left to fill in the question marks (1, 4, 5, 7, 8). We will now simplify our possible options by using an * to denote the each of the five choices (*, *, *, *, *).
So to fill the first question mark we use one of the 5 choices:
2 * ? ?
Now we have 4 choices left (*, *, *, *).
Fill in the next question mark with one of the choices.
2 * * ?
Now we have 3 choices left (*, *, *).
Fill in the last question mark.
2 * * *
As you can see to we had 5 choices for the first digit, 4 choices for the second digit, and 3 choices for the last digit. Simply use the PRODUCT RULE to find your answer.
5 X 4 X 3 = 60
There are 60 numbers between 2000 and 3000 using 1, 2, 4, 5, 7, and 8 only once.
--------------------------------------------------------------------------------------------------------------------------------------------
(P A R T . 2)
3 is not one of the possible digits so we now ask ourselves, how many numbers are there between 4000 and 5000 (using 1, 2, 4, 5, 7, and 8 only once)? This requires you to solve this puzzle:
4 ? ? ?, where each question mark represents 1 of 6 possible digits
BUT this time lets use LOGIC to determine this answer.
4 is already used so that leaves only five digits (1, 2, 5, 7, and 8). Simplifying our process makes the 5 choices (*, *, *, *, *).
Hmmm...
Since there are only 5 choices again to fill in 3 question marks our answer will be the same as the answer in part 1. But so you get some practice go ahead and use the PRODUCT RULE again:
5 X 4 X 3 = 60
There are 60 numbers between 4000 and 5000 using 1, 2, 4, 5, 7, and 8 only once.
--------------------------------------------------------------------------------------------------------------------------------------------
(P A R T . 3)
Combining the answers from part 1 and part 2 yields:
60 + 60 = 120
Thus, there are a total of 120 numbers between 2000 and 5000 using 1, 2, 4, 5, 7, and 8 only once.
source(s):
Rosen, K. H. (2007). Discrete Mathematics and Its Applications, 6/e.
Rosen, K. H. (2007). Discrete Mathematics and Its Applications, 6/e.
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M$
First thing is to try to imagine what kind of numbers are these. You can even try randomly some numbers until you find a pattern.
Does 124 satisfy the conditions? No, because 124 is not between 2000 and 5000
What about 5781? No because it is bigger than 5000
What about 5543? No, because it contains the number 3 which is not on the list
Can we find a pattern of it?
We see that the number has 4 digits.
We see the number cannot start with 1,5,7 or 8
So the number starts with 2 or 4
We have then two cases: the numbers that start with 2 of the form 2*** and the numbers that start with 4 of the form 4****
Say we try to find the numbers of the form 2***
The unknown digits are from the list but exclude the 2, so they are from the digits 1,4,5,7 and 8
How many choices? 5 choices.
How many we want to select ? We want to select 3 numbers out of 5
Let's imagine some numbers
2145, 2187, 2578 and so on
You can enumerate them all
2145,2147,2148, 2154,2157,2158,2174,2175,2178,2184,2185,2187 (how many ?12)
and similar for those that start with 24,25,27 and 28
You will notice there's the same count for numbers that start with 21,24,25,27,28
That's the brute force way. It works, it is fine-a display of patience. Patience is good in mathematics.
However, we can use shortcuts, for those who learned about combinations and permutations
The combination of n taken by k is also called n choose k
Let's say we want to choose 2 persons out of three A,B and C such that the order does not matter
We can choose AB, AC, or BC
3 possibilities
But what if the order matter, such that AB is different than BA. For example, the mom will select a son to wash the dishes and the other to bring the newspaper. The order would matter
How many choices are? AB,BA,BC,CB,AC,CA
6 possibilities
So in the first case we have combination of 3 taken by 2 (or 3 choose 2)
In the second case we have permutation of 3 taken by 2
In general, there are formulas for them
n choose k is equal with n!/(k!*(n-k)!)
while permutation of n taken by k is n!/(n-k)!
where n! is the factorial n!=1*2*...*n
Let's try to use these formulas for our example
3 choose 2=3!/2!*1!*=6/2=3
permutation of 3 taken by 2 is 3!/1!=6/1=6
It works
Let's come back to our problem
We had to choose 3 out of 5 digits
Does the order matter?
What formulas do we use?
What is the total number? Attention-you have to count both 2*** and 4*** numbers
If you have further questions, feel free to ask
Does 124 satisfy the conditions? No, because 124 is not between 2000 and 5000
What about 5781? No because it is bigger than 5000
What about 5543? No, because it contains the number 3 which is not on the list
Can we find a pattern of it?
We see that the number has 4 digits.
We see the number cannot start with 1,5,7 or 8
So the number starts with 2 or 4
We have then two cases: the numbers that start with 2 of the form 2*** and the numbers that start with 4 of the form 4****
Say we try to find the numbers of the form 2***
The unknown digits are from the list but exclude the 2, so they are from the digits 1,4,5,7 and 8
How many choices? 5 choices.
How many we want to select ? We want to select 3 numbers out of 5
Let's imagine some numbers
2145, 2187, 2578 and so on
You can enumerate them all
2145,2147,2148, 2154,2157,2158,2174,2175,2178,2184,2185,2187 (how many ?12)
and similar for those that start with 24,25,27 and 28
You will notice there's the same count for numbers that start with 21,24,25,27,28
That's the brute force way. It works, it is fine-a display of patience. Patience is good in mathematics.
However, we can use shortcuts, for those who learned about combinations and permutations
The combination of n taken by k is also called n choose k
Let's say we want to choose 2 persons out of three A,B and C such that the order does not matter
We can choose AB, AC, or BC
3 possibilities
But what if the order matter, such that AB is different than BA. For example, the mom will select a son to wash the dishes and the other to bring the newspaper. The order would matter
How many choices are? AB,BA,BC,CB,AC,CA
6 possibilities
So in the first case we have combination of 3 taken by 2 (or 3 choose 2)
In the second case we have permutation of 3 taken by 2
In general, there are formulas for them
n choose k is equal with n!/(k!*(n-k)!)
while permutation of n taken by k is n!/(n-k)!
where n! is the factorial n!=1*2*...*n
Let's try to use these formulas for our example
3 choose 2=3!/2!*1!*=6/2=3
permutation of 3 taken by 2 is 3!/1!=6/1=6
It works
Let's come back to our problem
We had to choose 3 out of 5 digits
Does the order matter?
What formulas do we use?
What is the total number? Attention-you have to count both 2*** and 4*** numbers
If you have further questions, feel free to ask
source(s):
my mind
my mind
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M$
OK everyone, relax. Yes, it's common courtesy to look at the other people's solutions before posting your own. But the length of the answer of @13rand0n makes it unclear that it's the same solution. It's also not right to attack another user for giving a similar answer.
That's the thing, your answer was long and overly complicated. The only reason I answered is because I saw how long the other two answers were and thought it needed to be explained in a simpler way. It was really easy for me to write a simple answer because the answer IS simple, and I could tell that neither of you communicated that by looking at how long your answers were.
I am an engineering student just like you except that I am a year ahead of you. Doesn't it make sense that I would know the answer to this question as soon as I read it?
@13rand0n I actually didn't read your answer at all before I did this. It's a pretty easy problem, you shouldn't assume that just because someone else was capable of working the problem correctly that they copied you.
My answer is intentionally complicated. I wanted to give the students more than a straightforward answer. Some students give formulas without having a feeling of what the particular cases are(in this case, numbers). I wished the students to give some concrete examples before applying the formulas. That's my personal opinion.
Of course your solution is great, it's how the solutions should be posted in the final stage. But before that, there's a lot of thinking "behind the scenes" going on.
It's good you posted the solution-it fills a gap. In fact, it's how the formula for permutations was deduced.
@13rand0n
Yes I'm sure you didn't apply anything you saw on this page before writing your answer. How easy it must have been to write a not-so "overly complicated" answer without looking at the other methods. If a method was already shown using the product rule, why are you going to repeat that answer? You applied the exact same method I did. It is common courtesy to not repeat an answer if it is already on a page.