Mixing
From digits 0, 1, 2, 3, 4, 5, 6, how many four-digit even numbers with distinct digits can be constructed?
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Trying to iron out the kinks from my thought process. I want to start backwards at the fourth digit. This gives me four options: $0,2,4,6$. So, _ * _ * _ * $4$.
Here's where I might be getting confused. Now I go back to the front to select the first digit. I can pick from six options which are $1,2,3,4,5,6$. However, does this leave me with six options if a $0$ is chosen for digit four, or five options if anything other than $0$ is chosen for digit four?
If this is an unsafe thought process, then call me out.
combinatorics
asked 1 hour ago
Ludwigthestud
512
add a comment |Â
up vote
3
down vote
favorite
Trying to iron out the kinks from my thought process. I want to start backwards at the fourth digit. This gives me four options: $0,2,4,6$. So, _ * _ * _ * $4$.
Here's where I might be getting confused. Now I go back to the front to select the first digit. I can pick from six options which are $1,2,3,4,5,6$. However, does this leave me with six options if a $0$ is chosen for digit four, or five options if anything other than $0$ is chosen for digit four?
If this is an unsafe thought process, then call me out.
combinatorics
asked 1 hour ago
Ludwigthestud
512
2
That idea is correct, you have to consider both cases and add the number of possibilities for both together to get the answer
– PhysMath
1 hour ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Trying to iron out the kinks from my thought process. I want to start backwards at the fourth digit. This gives me four options: $0,2,4,6$. So, _ * _ * _ * $4$.
Here's where I might be getting confused. Now I go back to the front to select the first digit. I can pick from six options which are $1,2,3,4,5,6$. However, does this leave me with six options if a $0$ is chosen for digit four, or five options if anything other than $0$ is chosen for digit four?
If this is an unsafe thought process, then call me out.
combinatorics
asked 1 hour ago
Ludwigthestud
512
Trying to iron out the kinks from my thought process. I want to start backwards at the fourth digit. This gives me four options: $0,2,4,6$. So, _ * _ * _ * $4$.
Here's where I might be getting confused. Now I go back to the front to select the first digit. I can pick from six options which are $1,2,3,4,5,6$. However, does this leave me with six options if a $0$ is chosen for digit four, or five options if anything other than $0$ is chosen for digit four?
If this is an unsafe thought process, then call me out.
combinatorics
combinatorics
asked 1 hour ago
Ludwigthestud
512
asked 1 hour ago
Ludwigthestud
512
asked 1 hour ago
Ludwigthestud
512
asked 1 hour ago
Ludwigthestud
512
asked 1 hour ago
Ludwigthestud
512
512
2
That idea is correct, you have to consider both cases and add the number of possibilities for both together to get the answer
– PhysMath
1 hour ago
add a comment |Â
2
That idea is correct, you have to consider both cases and add the number of possibilities for both together to get the answer
– PhysMath
1 hour ago
2
2
That idea is correct, you have to consider both cases and add the number of possibilities for both together to get the answer
– PhysMath
1 hour ago
That idea is correct, you have to consider both cases and add the number of possibilities for both together to get the answer
– PhysMath
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
You consider two cases, ending in $0$ or ending in $2,4,6$.
Suppose the last digit is $0$, then as you say, you have $6$ options for the first number, $5$ for the second, and $4$ for the third. Thus, there are $120$ distinct numbers that end in $0$.
Now suppose you pick a last digit that is not $0$. There are $3$ options for this. Then, you have only $5$ numbers to pick from for the first digit. However, you still have $0$ as an option for the second and third digits. In other words, there are also $5$ options for the second digit and $4$ options for the third. So there are $300$ options in this case.
Thus, there are a total $120 + 300 = 420$ numbers with the restrictions you gave.
answered 1 hour ago
PhysMath
1059
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PhysMath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
2
down vote
Your ideas are good.
You might distinct into two cases.
When the last digit is $0$ and when it is not $0$.
Since the number is supposed to be an even number with four digits, the last number has to be even and the first digit can not be $0$.
If the last number is $0$, there are $6cdot 5cdot 4$ possibilities for the other three digits.
If the last number is not $0$, the first number can not be $0$. This leaves $5$ possibilites for the first digit. Since the $0$ is illegal.
And $3$ for the last digit. For the other two we have $5cdot 4$ possibilities to choose. Since we have to again consider the $0$.
Hence $5cdot 5cdot 4cdot 3$ possibilites over all.
We have to add these to get the number of possible ways to construct such a number, which should be:
$5cdot 5cdot 4cdot 3+6cdot 5cdot 4$
answered 1 hour ago
Cornman
2,82921228
I would stress the fact that the number is even means we select the units digit first to ensure it is even, then pick the thousands digit since we still have to deal with the restriction that the leading digit cannot be zero.
– N. F. Taussig
54 mins ago
Yes, I meant it like that. But I got lost in counting which lead to editing my post several times.
– Cornman
52 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
You consider two cases, ending in $0$ or ending in $2,4,6$.
Suppose the last digit is $0$, then as you say, you have $6$ options for the first number, $5$ for the second, and $4$ for the third. Thus, there are $120$ distinct numbers that end in $0$.
Now suppose you pick a last digit that is not $0$. There are $3$ options for this. Then, you have only $5$ numbers to pick from for the first digit. However, you still have $0$ as an option for the second and third digits. In other words, there are also $5$ options for the second digit and $4$ options for the third. So there are $300$ options in this case.
Thus, there are a total $120 + 300 = 420$ numbers with the restrictions you gave.
answered 1 hour ago
PhysMath
1059
New contributor
PhysMath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
You consider two cases, ending in $0$ or ending in $2,4,6$.
Suppose the last digit is $0$, then as you say, you have $6$ options for the first number, $5$ for the second, and $4$ for the third. Thus, there are $120$ distinct numbers that end in $0$.
Now suppose you pick a last digit that is not $0$. There are $3$ options for this. Then, you have only $5$ numbers to pick from for the first digit. However, you still have $0$ as an option for the second and third digits. In other words, there are also $5$ options for the second digit and $4$ options for the third. So there are $300$ options in this case.
Thus, there are a total $120 + 300 = 420$ numbers with the restrictions you gave.
answered 1 hour ago
PhysMath
1059
New contributor
PhysMath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You consider two cases, ending in $0$ or ending in $2,4,6$.
Suppose the last digit is $0$, then as you say, you have $6$ options for the first number, $5$ for the second, and $4$ for the third. Thus, there are $120$ distinct numbers that end in $0$.
Now suppose you pick a last digit that is not $0$. There are $3$ options for this. Then, you have only $5$ numbers to pick from for the first digit. However, you still have $0$ as an option for the second and third digits. In other words, there are also $5$ options for the second digit and $4$ options for the third. So there are $300$ options in this case.
Thus, there are a total $120 + 300 = 420$ numbers with the restrictions you gave.
answered 1 hour ago
PhysMath
1059
New contributor
PhysMath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You consider two cases, ending in $0$ or ending in $2,4,6$.
Suppose the last digit is $0$, then as you say, you have $6$ options for the first number, $5$ for the second, and $4$ for the third. Thus, there are $120$ distinct numbers that end in $0$.
Now suppose you pick a last digit that is not $0$. There are $3$ options for this. Then, you have only $5$ numbers to pick from for the first digit. However, you still have $0$ as an option for the second and third digits. In other words, there are also $5$ options for the second digit and $4$ options for the third. So there are $300$ options in this case.
Thus, there are a total $120 + 300 = 420$ numbers with the restrictions you gave.
answered 1 hour ago
PhysMath
1059
New contributor
PhysMath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
PhysMath
1059
New contributor
PhysMath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
PhysMath
1059
answered 1 hour ago
PhysMath
1059
1059
New contributor
PhysMath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
PhysMath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
PhysMath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
up vote
2
down vote
Your ideas are good.
You might distinct into two cases.
When the last digit is $0$ and when it is not $0$.
Since the number is supposed to be an even number with four digits, the last number has to be even and the first digit can not be $0$.
If the last number is $0$, there are $6cdot 5cdot 4$ possibilities for the other three digits.
If the last number is not $0$, the first number can not be $0$. This leaves $5$ possibilites for the first digit. Since the $0$ is illegal.
And $3$ for the last digit. For the other two we have $5cdot 4$ possibilities to choose. Since we have to again consider the $0$.
Hence $5cdot 5cdot 4cdot 3$ possibilites over all.
We have to add these to get the number of possible ways to construct such a number, which should be:
$5cdot 5cdot 4cdot 3+6cdot 5cdot 4$
answered 1 hour ago
Cornman
2,82921228
I would stress the fact that the number is even means we select the units digit first to ensure it is even, then pick the thousands digit since we still have to deal with the restriction that the leading digit cannot be zero.
– N. F. Taussig
54 mins ago
Yes, I meant it like that. But I got lost in counting which lead to editing my post several times.
– Cornman
52 mins ago
add a comment |Â
up vote
2
down vote
Your ideas are good.
You might distinct into two cases.
When the last digit is $0$ and when it is not $0$.
Since the number is supposed to be an even number with four digits, the last number has to be even and the first digit can not be $0$.
If the last number is $0$, there are $6cdot 5cdot 4$ possibilities for the other three digits.
If the last number is not $0$, the first number can not be $0$. This leaves $5$ possibilites for the first digit. Since the $0$ is illegal.
And $3$ for the last digit. For the other two we have $5cdot 4$ possibilities to choose. Since we have to again consider the $0$.
Hence $5cdot 5cdot 4cdot 3$ possibilites over all.
We have to add these to get the number of possible ways to construct such a number, which should be:
$5cdot 5cdot 4cdot 3+6cdot 5cdot 4$
answered 1 hour ago
Cornman
2,82921228
I would stress the fact that the number is even means we select the units digit first to ensure it is even, then pick the thousands digit since we still have to deal with the restriction that the leading digit cannot be zero.
– N. F. Taussig
54 mins ago
Yes, I meant it like that. But I got lost in counting which lead to editing my post several times.
– Cornman
52 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your ideas are good.
You might distinct into two cases.
When the last digit is $0$ and when it is not $0$.
Since the number is supposed to be an even number with four digits, the last number has to be even and the first digit can not be $0$.
If the last number is $0$, there are $6cdot 5cdot 4$ possibilities for the other three digits.
If the last number is not $0$, the first number can not be $0$. This leaves $5$ possibilites for the first digit. Since the $0$ is illegal.
And $3$ for the last digit. For the other two we have $5cdot 4$ possibilities to choose. Since we have to again consider the $0$.
Hence $5cdot 5cdot 4cdot 3$ possibilites over all.
We have to add these to get the number of possible ways to construct such a number, which should be:
$5cdot 5cdot 4cdot 3+6cdot 5cdot 4$
answered 1 hour ago
Cornman
2,82921228
Your ideas are good.
You might distinct into two cases.
When the last digit is $0$ and when it is not $0$.
Since the number is supposed to be an even number with four digits, the last number has to be even and the first digit can not be $0$.
If the last number is $0$, there are $6cdot 5cdot 4$ possibilities for the other three digits.
If the last number is not $0$, the first number can not be $0$. This leaves $5$ possibilites for the first digit. Since the $0$ is illegal.
And $3$ for the last digit. For the other two we have $5cdot 4$ possibilities to choose. Since we have to again consider the $0$.
Hence $5cdot 5cdot 4cdot 3$ possibilites over all.
We have to add these to get the number of possible ways to construct such a number, which should be:
$5cdot 5cdot 4cdot 3+6cdot 5cdot 4$
answered 1 hour ago
Cornman
2,82921228
edited 1 hour ago
answered 1 hour ago
Cornman
2,82921228
answered 1 hour ago
Cornman
2,82921228
answered 1 hour ago
Cornman
2,82921228
2,82921228
I would stress the fact that the number is even means we select the units digit first to ensure it is even, then pick the thousands digit since we still have to deal with the restriction that the leading digit cannot be zero.
– N. F. Taussig
54 mins ago
Yes, I meant it like that. But I got lost in counting which lead to editing my post several times.
– Cornman
52 mins ago
add a comment |Â
I would stress the fact that the number is even means we select the units digit first to ensure it is even, then pick the thousands digit since we still have to deal with the restriction that the leading digit cannot be zero.
– N. F. Taussig
54 mins ago
Yes, I meant it like that. But I got lost in counting which lead to editing my post several times.
– Cornman
52 mins ago
I would stress the fact that the number is even means we select the units digit first to ensure it is even, then pick the thousands digit since we still have to deal with the restriction that the leading digit cannot be zero.
– N. F. Taussig
54 mins ago
I would stress the fact that the number is even means we select the units digit first to ensure it is even, then pick the thousands digit since we still have to deal with the restriction that the leading digit cannot be zero.
– N. F. Taussig
54 mins ago
Yes, I meant it like that. But I got lost in counting which lead to editing my post several times.
– Cornman
52 mins ago
Yes, I meant it like that. But I got lost in counting which lead to editing my post several times.
– Cornman
52 mins ago
add a comment |Â
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2
That idea is correct, you have to consider both cases and add the number of possibilities for both together to get the answer
– PhysMath
1 hour ago