Mixing
Employee of the month
$begingroup$
How many ways can 9 students become employee of the month for the next three months?
My solution is simply 9×9×9=729. Just asking if it's correct though.
combinatorics
asked Feb 3 at 11:13
Isaiah LeobreraIsaiah Leobrera
281
$endgroup$
add a comment |
$begingroup$
How many ways can 9 students become employee of the month for the next three months?
My solution is simply 9×9×9=729. Just asking if it's correct though.
combinatorics
asked Feb 3 at 11:13
Isaiah LeobreraIsaiah Leobrera
281
$endgroup$
$begingroup$
You should explain why you’re multiplying 9 by itself twice.
$endgroup$
– Kyky
Feb 3 at 11:14
$begingroup$
If a student can be chosen multiple times, your solution is correct.
$endgroup$
– Peter
Feb 3 at 11:17
add a comment |
$begingroup$
How many ways can 9 students become employee of the month for the next three months?
My solution is simply 9×9×9=729. Just asking if it's correct though.
combinatorics
asked Feb 3 at 11:13
Isaiah LeobreraIsaiah Leobrera
281
$endgroup$
How many ways can 9 students become employee of the month for the next three months?
My solution is simply 9×9×9=729. Just asking if it's correct though.
combinatorics
combinatorics
asked Feb 3 at 11:13
Isaiah LeobreraIsaiah Leobrera
281
asked Feb 3 at 11:13
Isaiah LeobreraIsaiah Leobrera
281
asked Feb 3 at 11:13
Isaiah LeobreraIsaiah Leobrera
281
asked Feb 3 at 11:13
Isaiah LeobreraIsaiah Leobrera
281
asked Feb 3 at 11:13
Isaiah LeobreraIsaiah Leobrera
281
281
$begingroup$
You should explain why you’re multiplying 9 by itself twice.
$endgroup$
– Kyky
Feb 3 at 11:14
$begingroup$
If a student can be chosen multiple times, your solution is correct.
$endgroup$
– Peter
Feb 3 at 11:17
add a comment |
$begingroup$
You should explain why you’re multiplying 9 by itself twice.
$endgroup$
– Kyky
Feb 3 at 11:14
$begingroup$
If a student can be chosen multiple times, your solution is correct.
$endgroup$
– Peter
Feb 3 at 11:17
$begingroup$
You should explain why you’re multiplying 9 by itself twice.
$endgroup$
– Kyky
Feb 3 at 11:14
$begingroup$
You should explain why you’re multiplying 9 by itself twice.
$endgroup$
– Kyky
Feb 3 at 11:14
$begingroup$
If a student can be chosen multiple times, your solution is correct.
$endgroup$
– Peter
Feb 3 at 11:17
$begingroup$
If a student can be chosen multiple times, your solution is correct.
$endgroup$
– Peter
Feb 3 at 11:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If a student can be an employee multiple times, then it's correct.
Otherwise, the answer is simply $9cdot8cdot7=504$ since there are $8$ choices on the second month, $7$ on the third month.
answered Feb 3 at 11:19
abc...abc...
3,242739
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
If a student can be an employee multiple times, then it's correct.
Otherwise, the answer is simply $9cdot8cdot7=504$ since there are $8$ choices on the second month, $7$ on the third month.
answered Feb 3 at 11:19
abc...abc...
3,242739
$endgroup$
add a comment |
$begingroup$
If a student can be an employee multiple times, then it's correct.
Otherwise, the answer is simply $9cdot8cdot7=504$ since there are $8$ choices on the second month, $7$ on the third month.
answered Feb 3 at 11:19
abc...abc...
3,242739
$endgroup$
add a comment |
$begingroup$
If a student can be an employee multiple times, then it's correct.
Otherwise, the answer is simply $9cdot8cdot7=504$ since there are $8$ choices on the second month, $7$ on the third month.
answered Feb 3 at 11:19
abc...abc...
3,242739
$endgroup$
If a student can be an employee multiple times, then it's correct.
Otherwise, the answer is simply $9cdot8cdot7=504$ since there are $8$ choices on the second month, $7$ on the third month.
answered Feb 3 at 11:19
abc...abc...
3,242739
answered Feb 3 at 11:19
abc...abc...
3,242739
answered Feb 3 at 11:19
abc...abc...
3,242739
answered Feb 3 at 11:19
abc...abc...
3,242739
3,242739
add a comment |
add a comment |
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$begingroup$
You should explain why you’re multiplying 9 by itself twice.
$endgroup$
– Kyky
Feb 3 at 11:14
$begingroup$
If a student can be chosen multiple times, your solution is correct.
$endgroup$
– Peter
Feb 3 at 11:17