Combinatorics arranging repeated numbers
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In the multiset $B=${$5,5,5,7,9,11$}, ¿How many samples of size $3$, without order and without replacement can be extracted of the population?
I think the answer is $8$, but i don't know how to use the formula of combination, because the $5$ is in the set three times. Any hints?
combinatorics combinations
asked Nov 17 at 23:51
Rodrigo Pizarro
834217
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up vote
1
down vote
favorite
In the multiset $B=${$5,5,5,7,9,11$}, ¿How many samples of size $3$, without order and without replacement can be extracted of the population?
I think the answer is $8$, but i don't know how to use the formula of combination, because the $5$ is in the set three times. Any hints?
combinatorics combinations
asked Nov 17 at 23:51
Rodrigo Pizarro
834217
1
modified, sorry.
– Rodrigo Pizarro
Nov 17 at 23:59
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In the multiset $B=${$5,5,5,7,9,11$}, ¿How many samples of size $3$, without order and without replacement can be extracted of the population?
I think the answer is $8$, but i don't know how to use the formula of combination, because the $5$ is in the set three times. Any hints?
combinatorics combinations
asked Nov 17 at 23:51
Rodrigo Pizarro
834217
In the multiset $B=${$5,5,5,7,9,11$}, ¿How many samples of size $3$, without order and without replacement can be extracted of the population?
I think the answer is $8$, but i don't know how to use the formula of combination, because the $5$ is in the set three times. Any hints?
combinatorics combinations
combinatorics combinations
asked Nov 17 at 23:51
Rodrigo Pizarro
834217
asked Nov 17 at 23:51
Rodrigo Pizarro
834217
edited Nov 17 at 23:59
asked Nov 17 at 23:51
Rodrigo Pizarro
834217
asked Nov 17 at 23:51
Rodrigo Pizarro
834217
asked Nov 17 at 23:51
Rodrigo Pizarro
834217
834217
1
modified, sorry.
– Rodrigo Pizarro
Nov 17 at 23:59
add a comment |
1
modified, sorry.
– Rodrigo Pizarro
Nov 17 at 23:59
1
1
modified, sorry.
– Rodrigo Pizarro
Nov 17 at 23:59
modified, sorry.
– Rodrigo Pizarro
Nov 17 at 23:59
add a comment |
1 Answer
1
active
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votes
up vote
2
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accepted
Your answer is correct.
Strategy: Consider cases, depending on the number of fives used.
- No fives are used: Choose three of the other three numbers in $B$.
- One five is used: Choose two of the other three numbers in $B$.
- Two fives are used: Choose one of the other three numbers in $B$.
- Three fives are used: Choose none of the other three numbers in $B$.
Since the cases are mutually exclusive and exhaustive, add the results to get the total.
answered Nov 18 at 0:03
N. F. Taussig
42.7k93254
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your answer is correct.
Strategy: Consider cases, depending on the number of fives used.
- No fives are used: Choose three of the other three numbers in $B$.
- One five is used: Choose two of the other three numbers in $B$.
- Two fives are used: Choose one of the other three numbers in $B$.
- Three fives are used: Choose none of the other three numbers in $B$.
Since the cases are mutually exclusive and exhaustive, add the results to get the total.
answered Nov 18 at 0:03
N. F. Taussig
42.7k93254
add a comment |
up vote
2
down vote
accepted
Your answer is correct.
Strategy: Consider cases, depending on the number of fives used.
- No fives are used: Choose three of the other three numbers in $B$.
- One five is used: Choose two of the other three numbers in $B$.
- Two fives are used: Choose one of the other three numbers in $B$.
- Three fives are used: Choose none of the other three numbers in $B$.
Since the cases are mutually exclusive and exhaustive, add the results to get the total.
answered Nov 18 at 0:03
N. F. Taussig
42.7k93254
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your answer is correct.
Strategy: Consider cases, depending on the number of fives used.
- No fives are used: Choose three of the other three numbers in $B$.
- One five is used: Choose two of the other three numbers in $B$.
- Two fives are used: Choose one of the other three numbers in $B$.
- Three fives are used: Choose none of the other three numbers in $B$.
Since the cases are mutually exclusive and exhaustive, add the results to get the total.
answered Nov 18 at 0:03
N. F. Taussig
42.7k93254
Your answer is correct.
Strategy: Consider cases, depending on the number of fives used.
- No fives are used: Choose three of the other three numbers in $B$.
- One five is used: Choose two of the other three numbers in $B$.
- Two fives are used: Choose one of the other three numbers in $B$.
- Three fives are used: Choose none of the other three numbers in $B$.
Since the cases are mutually exclusive and exhaustive, add the results to get the total.
answered Nov 18 at 0:03
N. F. Taussig
42.7k93254
answered Nov 18 at 0:03
N. F. Taussig
42.7k93254
answered Nov 18 at 0:03
N. F. Taussig
42.7k93254
answered Nov 18 at 0:03
N. F. Taussig
42.7k93254
42.7k93254
add a comment |
add a comment |
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modified, sorry.
– Rodrigo Pizarro
Nov 17 at 23:59