Number of ways to form 3 digit numbers from 1,2,5,6,8 if each digit can only be used once. [closed]
Number of ways to form 3 digit even numbers from 1,2,5,6,9 if each digit can only be used once.
my understanding is that for even numbers, the last digit must be even or 0.
picking 3 numbers ->
my last pick (3rd number) i understand that i only need to have 2 available cards to pick and these 2 cards needs to be an even number (2,6)
so working backwards, my first pick will have 4 numbers to choose from,
So $4 cdot 3 cdot 2 $ is the answer.
But having 5 numbers to choose from, why is the first pick only 4 different possible ways ?
combinatorics
asked Nov 24 at 17:50
mutu mumu
374
closed as unclear what you're asking by amWhy, lulu, Joel Reyes Noche, Brahadeesh, Paul Frost Nov 25 at 15:49
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. 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.
add a comment |
Number of ways to form 3 digit even numbers from 1,2,5,6,9 if each digit can only be used once.
my understanding is that for even numbers, the last digit must be even or 0.
picking 3 numbers ->
my last pick (3rd number) i understand that i only need to have 2 available cards to pick and these 2 cards needs to be an even number (2,6)
so working backwards, my first pick will have 4 numbers to choose from,
So $4 cdot 3 cdot 2 $ is the answer.
But having 5 numbers to choose from, why is the first pick only 4 different possible ways ?
combinatorics
asked Nov 24 at 17:50
mutu mumu
374
closed as unclear what you're asking by amWhy, lulu, Joel Reyes Noche, Brahadeesh, Paul Frost Nov 25 at 15:49
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. 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.
2
This doesn't make much sense to me. You don't say anything about even numbers in the problem statement, but then you say the last digit must be even. Please edit the question to clarify it.
– saulspatz
Nov 24 at 17:54
1
This question is very poorly written. First, as has been pointed out, you never say that you are only interested in even numbers though your calculation assumes this. Secondly, the list is ${1,2,5,6,8}$ in the header and ${1,2,5,6,9}$ in the body which makes a big difference if you really meant to focus on even numbers. Please edit for clarity.
– lulu
Nov 24 at 18:11
Voting to close the question as it is not clear what you are asking.
– lulu
Nov 24 at 18:19
add a comment |
Number of ways to form 3 digit even numbers from 1,2,5,6,9 if each digit can only be used once.
my understanding is that for even numbers, the last digit must be even or 0.
picking 3 numbers ->
my last pick (3rd number) i understand that i only need to have 2 available cards to pick and these 2 cards needs to be an even number (2,6)
so working backwards, my first pick will have 4 numbers to choose from,
So $4 cdot 3 cdot 2 $ is the answer.
But having 5 numbers to choose from, why is the first pick only 4 different possible ways ?
combinatorics
asked Nov 24 at 17:50
mutu mumu
374
Number of ways to form 3 digit even numbers from 1,2,5,6,9 if each digit can only be used once.
my understanding is that for even numbers, the last digit must be even or 0.
picking 3 numbers ->
my last pick (3rd number) i understand that i only need to have 2 available cards to pick and these 2 cards needs to be an even number (2,6)
so working backwards, my first pick will have 4 numbers to choose from,
So $4 cdot 3 cdot 2 $ is the answer.
But having 5 numbers to choose from, why is the first pick only 4 different possible ways ?
combinatorics
combinatorics
asked Nov 24 at 17:50
mutu mumu
374
asked Nov 24 at 17:50
mutu mumu
374
edited Nov 24 at 18:13
asked Nov 24 at 17:50
mutu mumu
374
asked Nov 24 at 17:50
mutu mumu
374
asked Nov 24 at 17:50
mutu mumu
374
374
closed as unclear what you're asking by amWhy, lulu, Joel Reyes Noche, Brahadeesh, Paul Frost Nov 25 at 15:49
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. 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.
closed as unclear what you're asking by amWhy, lulu, Joel Reyes Noche, Brahadeesh, Paul Frost Nov 25 at 15:49
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. 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.
2
This doesn't make much sense to me. You don't say anything about even numbers in the problem statement, but then you say the last digit must be even. Please edit the question to clarify it.
– saulspatz
Nov 24 at 17:54
1
This question is very poorly written. First, as has been pointed out, you never say that you are only interested in even numbers though your calculation assumes this. Secondly, the list is ${1,2,5,6,8}$ in the header and ${1,2,5,6,9}$ in the body which makes a big difference if you really meant to focus on even numbers. Please edit for clarity.
– lulu
Nov 24 at 18:11
Voting to close the question as it is not clear what you are asking.
– lulu
Nov 24 at 18:19
add a comment |
2
This doesn't make much sense to me. You don't say anything about even numbers in the problem statement, but then you say the last digit must be even. Please edit the question to clarify it.
– saulspatz
Nov 24 at 17:54
1
This question is very poorly written. First, as has been pointed out, you never say that you are only interested in even numbers though your calculation assumes this. Secondly, the list is ${1,2,5,6,8}$ in the header and ${1,2,5,6,9}$ in the body which makes a big difference if you really meant to focus on even numbers. Please edit for clarity.
– lulu
Nov 24 at 18:11
Voting to close the question as it is not clear what you are asking.
– lulu
Nov 24 at 18:19
2
2
This doesn't make much sense to me. You don't say anything about even numbers in the problem statement, but then you say the last digit must be even. Please edit the question to clarify it.
– saulspatz
Nov 24 at 17:54
This doesn't make much sense to me. You don't say anything about even numbers in the problem statement, but then you say the last digit must be even. Please edit the question to clarify it.
– saulspatz
Nov 24 at 17:54
1
1
This question is very poorly written. First, as has been pointed out, you never say that you are only interested in even numbers though your calculation assumes this. Secondly, the list is ${1,2,5,6,8}$ in the header and ${1,2,5,6,9}$ in the body which makes a big difference if you really meant to focus on even numbers. Please edit for clarity.
– lulu
Nov 24 at 18:11
This question is very poorly written. First, as has been pointed out, you never say that you are only interested in even numbers though your calculation assumes this. Secondly, the list is ${1,2,5,6,8}$ in the header and ${1,2,5,6,9}$ in the body which makes a big difference if you really meant to focus on even numbers. Please edit for clarity.
– lulu
Nov 24 at 18:11
Voting to close the question as it is not clear what you are asking.
– lulu
Nov 24 at 18:19
Voting to close the question as it is not clear what you are asking.
– lulu
Nov 24 at 18:19
add a comment |
3 Answers
3
active
oldest
votes
But having 5 numbers to choose from, why is the first pick only 4 different possible ways ?
Because there aren't five possible numbers to choose from. One of them you have already used in the last position.
answered Nov 24 at 17:53
user3482749
2,421414
add a comment |
You started out correctly by first focusing on the last digit, for which there are indeed $3$ options. So, after you have picked one of those, there are $4$ digits left to pick from for the first digit, and then there are $3$ choices left for the second digit. So: $3cdot 4 cdot3$ possibilities
So, the correct answer is not $4cdot 3 cdot 2$
Moreover, the choice for the first digit is not $5$ if you first pick one of the $3$ even numbers for the last digit first.
Now, you could start by picking one of the $5$ Numbers for the first digit first, but the issue is that if you do that, you eill need to distinguish between the case where you picked an even number for that first digit or an odd number, and now you get a more complicated equation
answered Nov 24 at 18:22
Bram28
60.1k44590
add a comment |
Pick the last digit first. You have $3$ choices ($2,6$ or $8$).
Then pick the first digit. As you have already picked a digit you have only $4$ choices remaining.
Then pick the second digit. As you have already picked two you have only three choices remaining.
So the number of ways to do this is $3*4*3 = 36$
If you don't pick the last digit first you will have $5$ choices for the first choice and $4$ for the second but when coming to choosing the last digit you do not know if you have $3$ choices, $2$ chooses or only $1$. So you can't do it that way without tricky subcases (which can be done but is trickier). ANd it doesn't matter which order you choose the digits. So it's a good idea to pick the most restrictive digit first.
Maybe the following make it clearer.
$overbrace{begin{cases}overbrace{begin{cases} \
overbrace{begin{cases} 152\162\182
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 512\562\582
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 612\652\682
end{cases}}^{text{first digit is 6 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 812\852\862
end{cases}}^{text{first digit is 8 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 2 (there are 4 choices for the first digit)}}\
overbrace{begin{cases} \
overbrace{begin{cases} 156\126\186
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 216\256\286
end{cases}}^{text{first digit is 2 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 516\526\586
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 816\826\856
end{cases}}^{text{first digit is 8 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 6 (there are 4 choices for the first digit)}}\
overbrace{begin{cases} \
overbrace{begin{cases} 128\158\168
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 218\258\268
end{cases}}^{text{first digit is 2 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 518\528\568
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 618\628\658
end{cases}}^{text{first digit is 6 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 8 (there are 4 choices for the first digit)}}end{cases}}^{text{There are 3 choices for the last digit; either 2,6, or 8}}$
answered Nov 24 at 18:49
fleablood
68.1k22684
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
But having 5 numbers to choose from, why is the first pick only 4 different possible ways ?
Because there aren't five possible numbers to choose from. One of them you have already used in the last position.
answered Nov 24 at 17:53
user3482749
2,421414
add a comment |
But having 5 numbers to choose from, why is the first pick only 4 different possible ways ?
Because there aren't five possible numbers to choose from. One of them you have already used in the last position.
answered Nov 24 at 17:53
user3482749
2,421414
add a comment |
But having 5 numbers to choose from, why is the first pick only 4 different possible ways ?
Because there aren't five possible numbers to choose from. One of them you have already used in the last position.
answered Nov 24 at 17:53
user3482749
2,421414
But having 5 numbers to choose from, why is the first pick only 4 different possible ways ?
Because there aren't five possible numbers to choose from. One of them you have already used in the last position.
answered Nov 24 at 17:53
user3482749
2,421414
answered Nov 24 at 17:53
user3482749
2,421414
answered Nov 24 at 17:53
user3482749
2,421414
answered Nov 24 at 17:53
user3482749
2,421414
2,421414
add a comment |
add a comment |
You started out correctly by first focusing on the last digit, for which there are indeed $3$ options. So, after you have picked one of those, there are $4$ digits left to pick from for the first digit, and then there are $3$ choices left for the second digit. So: $3cdot 4 cdot3$ possibilities
So, the correct answer is not $4cdot 3 cdot 2$
Moreover, the choice for the first digit is not $5$ if you first pick one of the $3$ even numbers for the last digit first.
Now, you could start by picking one of the $5$ Numbers for the first digit first, but the issue is that if you do that, you eill need to distinguish between the case where you picked an even number for that first digit or an odd number, and now you get a more complicated equation
answered Nov 24 at 18:22
Bram28
60.1k44590
add a comment |
You started out correctly by first focusing on the last digit, for which there are indeed $3$ options. So, after you have picked one of those, there are $4$ digits left to pick from for the first digit, and then there are $3$ choices left for the second digit. So: $3cdot 4 cdot3$ possibilities
So, the correct answer is not $4cdot 3 cdot 2$
Moreover, the choice for the first digit is not $5$ if you first pick one of the $3$ even numbers for the last digit first.
Now, you could start by picking one of the $5$ Numbers for the first digit first, but the issue is that if you do that, you eill need to distinguish between the case where you picked an even number for that first digit or an odd number, and now you get a more complicated equation
answered Nov 24 at 18:22
Bram28
60.1k44590
add a comment |
You started out correctly by first focusing on the last digit, for which there are indeed $3$ options. So, after you have picked one of those, there are $4$ digits left to pick from for the first digit, and then there are $3$ choices left for the second digit. So: $3cdot 4 cdot3$ possibilities
So, the correct answer is not $4cdot 3 cdot 2$
Moreover, the choice for the first digit is not $5$ if you first pick one of the $3$ even numbers for the last digit first.
Now, you could start by picking one of the $5$ Numbers for the first digit first, but the issue is that if you do that, you eill need to distinguish between the case where you picked an even number for that first digit or an odd number, and now you get a more complicated equation
answered Nov 24 at 18:22
Bram28
60.1k44590
You started out correctly by first focusing on the last digit, for which there are indeed $3$ options. So, after you have picked one of those, there are $4$ digits left to pick from for the first digit, and then there are $3$ choices left for the second digit. So: $3cdot 4 cdot3$ possibilities
So, the correct answer is not $4cdot 3 cdot 2$
Moreover, the choice for the first digit is not $5$ if you first pick one of the $3$ even numbers for the last digit first.
Now, you could start by picking one of the $5$ Numbers for the first digit first, but the issue is that if you do that, you eill need to distinguish between the case where you picked an even number for that first digit or an odd number, and now you get a more complicated equation
answered Nov 24 at 18:22
Bram28
60.1k44590
edited Nov 24 at 18:29
answered Nov 24 at 18:22
Bram28
60.1k44590
answered Nov 24 at 18:22
Bram28
60.1k44590
answered Nov 24 at 18:22
Bram28
60.1k44590
60.1k44590
add a comment |
add a comment |
Pick the last digit first. You have $3$ choices ($2,6$ or $8$).
Then pick the first digit. As you have already picked a digit you have only $4$ choices remaining.
Then pick the second digit. As you have already picked two you have only three choices remaining.
So the number of ways to do this is $3*4*3 = 36$
If you don't pick the last digit first you will have $5$ choices for the first choice and $4$ for the second but when coming to choosing the last digit you do not know if you have $3$ choices, $2$ chooses or only $1$. So you can't do it that way without tricky subcases (which can be done but is trickier). ANd it doesn't matter which order you choose the digits. So it's a good idea to pick the most restrictive digit first.
Maybe the following make it clearer.
$overbrace{begin{cases}overbrace{begin{cases} \
overbrace{begin{cases} 152\162\182
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 512\562\582
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 612\652\682
end{cases}}^{text{first digit is 6 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 812\852\862
end{cases}}^{text{first digit is 8 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 2 (there are 4 choices for the first digit)}}\
overbrace{begin{cases} \
overbrace{begin{cases} 156\126\186
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 216\256\286
end{cases}}^{text{first digit is 2 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 516\526\586
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 816\826\856
end{cases}}^{text{first digit is 8 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 6 (there are 4 choices for the first digit)}}\
overbrace{begin{cases} \
overbrace{begin{cases} 128\158\168
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 218\258\268
end{cases}}^{text{first digit is 2 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 518\528\568
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 618\628\658
end{cases}}^{text{first digit is 6 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 8 (there are 4 choices for the first digit)}}end{cases}}^{text{There are 3 choices for the last digit; either 2,6, or 8}}$
answered Nov 24 at 18:49
fleablood
68.1k22684
add a comment |
Pick the last digit first. You have $3$ choices ($2,6$ or $8$).
Then pick the first digit. As you have already picked a digit you have only $4$ choices remaining.
Then pick the second digit. As you have already picked two you have only three choices remaining.
So the number of ways to do this is $3*4*3 = 36$
If you don't pick the last digit first you will have $5$ choices for the first choice and $4$ for the second but when coming to choosing the last digit you do not know if you have $3$ choices, $2$ chooses or only $1$. So you can't do it that way without tricky subcases (which can be done but is trickier). ANd it doesn't matter which order you choose the digits. So it's a good idea to pick the most restrictive digit first.
Maybe the following make it clearer.
$overbrace{begin{cases}overbrace{begin{cases} \
overbrace{begin{cases} 152\162\182
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 512\562\582
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 612\652\682
end{cases}}^{text{first digit is 6 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 812\852\862
end{cases}}^{text{first digit is 8 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 2 (there are 4 choices for the first digit)}}\
overbrace{begin{cases} \
overbrace{begin{cases} 156\126\186
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 216\256\286
end{cases}}^{text{first digit is 2 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 516\526\586
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 816\826\856
end{cases}}^{text{first digit is 8 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 6 (there are 4 choices for the first digit)}}\
overbrace{begin{cases} \
overbrace{begin{cases} 128\158\168
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 218\258\268
end{cases}}^{text{first digit is 2 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 518\528\568
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 618\628\658
end{cases}}^{text{first digit is 6 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 8 (there are 4 choices for the first digit)}}end{cases}}^{text{There are 3 choices for the last digit; either 2,6, or 8}}$
answered Nov 24 at 18:49
fleablood
68.1k22684
add a comment |
Pick the last digit first. You have $3$ choices ($2,6$ or $8$).
Then pick the first digit. As you have already picked a digit you have only $4$ choices remaining.
Then pick the second digit. As you have already picked two you have only three choices remaining.
So the number of ways to do this is $3*4*3 = 36$
If you don't pick the last digit first you will have $5$ choices for the first choice and $4$ for the second but when coming to choosing the last digit you do not know if you have $3$ choices, $2$ chooses or only $1$. So you can't do it that way without tricky subcases (which can be done but is trickier). ANd it doesn't matter which order you choose the digits. So it's a good idea to pick the most restrictive digit first.
Maybe the following make it clearer.
$overbrace{begin{cases}overbrace{begin{cases} \
overbrace{begin{cases} 152\162\182
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 512\562\582
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 612\652\682
end{cases}}^{text{first digit is 6 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 812\852\862
end{cases}}^{text{first digit is 8 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 2 (there are 4 choices for the first digit)}}\
overbrace{begin{cases} \
overbrace{begin{cases} 156\126\186
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 216\256\286
end{cases}}^{text{first digit is 2 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 516\526\586
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 816\826\856
end{cases}}^{text{first digit is 8 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 6 (there are 4 choices for the first digit)}}\
overbrace{begin{cases} \
overbrace{begin{cases} 128\158\168
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 218\258\268
end{cases}}^{text{first digit is 2 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 518\528\568
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 618\628\658
end{cases}}^{text{first digit is 6 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 8 (there are 4 choices for the first digit)}}end{cases}}^{text{There are 3 choices for the last digit; either 2,6, or 8}}$
answered Nov 24 at 18:49
fleablood
68.1k22684
Pick the last digit first. You have $3$ choices ($2,6$ or $8$).
Then pick the first digit. As you have already picked a digit you have only $4$ choices remaining.
Then pick the second digit. As you have already picked two you have only three choices remaining.
So the number of ways to do this is $3*4*3 = 36$
If you don't pick the last digit first you will have $5$ choices for the first choice and $4$ for the second but when coming to choosing the last digit you do not know if you have $3$ choices, $2$ chooses or only $1$. So you can't do it that way without tricky subcases (which can be done but is trickier). ANd it doesn't matter which order you choose the digits. So it's a good idea to pick the most restrictive digit first.
Maybe the following make it clearer.
$overbrace{begin{cases}overbrace{begin{cases} \
overbrace{begin{cases} 152\162\182
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 512\562\582
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 612\652\682
end{cases}}^{text{first digit is 6 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 812\852\862
end{cases}}^{text{first digit is 8 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 2 (there are 4 choices for the first digit)}}\
overbrace{begin{cases} \
overbrace{begin{cases} 156\126\186
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 216\256\286
end{cases}}^{text{first digit is 2 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 516\526\586
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 816\826\856
end{cases}}^{text{first digit is 8 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 6 (there are 4 choices for the first digit)}}\
overbrace{begin{cases} \
overbrace{begin{cases} 128\158\168
end{cases}}^{text{first digit is 1 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 218\258\268
end{cases}}^{text{first digit is 2 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 518\528\568
end{cases}}^{text{first digit is 5 (there are 3 choices for the second digit)}}\
overbrace{begin{cases} 618\628\658
end{cases}}^{text{first digit is 6 (there are 3 choices for the second digit)}}\
end{cases}}^{text{last digit is 8 (there are 4 choices for the first digit)}}end{cases}}^{text{There are 3 choices for the last digit; either 2,6, or 8}}$
answered Nov 24 at 18:49
fleablood
68.1k22684
edited Nov 24 at 18:58
answered Nov 24 at 18:49
fleablood
68.1k22684
answered Nov 24 at 18:49
fleablood
68.1k22684
answered Nov 24 at 18:49
fleablood
68.1k22684
68.1k22684
add a comment |
add a comment |
2
This doesn't make much sense to me. You don't say anything about even numbers in the problem statement, but then you say the last digit must be even. Please edit the question to clarify it.
– saulspatz
Nov 24 at 17:54
1
This question is very poorly written. First, as has been pointed out, you never say that you are only interested in even numbers though your calculation assumes this. Secondly, the list is ${1,2,5,6,8}$ in the header and ${1,2,5,6,9}$ in the body which makes a big difference if you really meant to focus on even numbers. Please edit for clarity.
– lulu
Nov 24 at 18:11
Voting to close the question as it is not clear what you are asking.
– lulu
Nov 24 at 18:19