Set and subsets
I understand most of the concept for set and subsets however, I do not understand the logic here.
Question : Consider the set S = {a,b,c,d,e,f,g,h,i,j}. How many 4-element subsets of S contain neither a nor b?
Solution: 4 elements to choose out of 8 = 70
I don't know where does the 4 comes from...
combinatorics discrete-mathematics
edited Nov 28 at 23:47
N. F. Taussig
43.5k93355
asked Nov 28 at 23:28
Laura1999
202
add a comment |
I understand most of the concept for set and subsets however, I do not understand the logic here.
Question : Consider the set S = {a,b,c,d,e,f,g,h,i,j}. How many 4-element subsets of S contain neither a nor b?
Solution: 4 elements to choose out of 8 = 70
I don't know where does the 4 comes from...
combinatorics discrete-mathematics
edited Nov 28 at 23:47
N. F. Taussig
43.5k93355
asked Nov 28 at 23:28
Laura1999
202
4
$4$ comes from the "..4-element subset .." and $8$ from $10$ elements ninus the $2$ excluded
– G Cab
Nov 28 at 23:31
Ho yeah sorry it's my mistake!
– Laura1999
Nov 28 at 23:39
Enumeration problems should be tagged combinatorics rather than probability.
– N. F. Taussig
Nov 28 at 23:47
add a comment |
I understand most of the concept for set and subsets however, I do not understand the logic here.
Question : Consider the set S = {a,b,c,d,e,f,g,h,i,j}. How many 4-element subsets of S contain neither a nor b?
Solution: 4 elements to choose out of 8 = 70
I don't know where does the 4 comes from...
combinatorics discrete-mathematics
edited Nov 28 at 23:47
N. F. Taussig
43.5k93355
asked Nov 28 at 23:28
Laura1999
202
I understand most of the concept for set and subsets however, I do not understand the logic here.
Question : Consider the set S = {a,b,c,d,e,f,g,h,i,j}. How many 4-element subsets of S contain neither a nor b?
Solution: 4 elements to choose out of 8 = 70
I don't know where does the 4 comes from...
combinatorics discrete-mathematics
combinatorics discrete-mathematics
edited Nov 28 at 23:47
N. F. Taussig
43.5k93355
asked Nov 28 at 23:28
Laura1999
202
edited Nov 28 at 23:47
N. F. Taussig
43.5k93355
asked Nov 28 at 23:28
Laura1999
202
edited Nov 28 at 23:47
N. F. Taussig
43.5k93355
edited Nov 28 at 23:47
N. F. Taussig
43.5k93355
edited Nov 28 at 23:47
N. F. Taussig
43.5k93355
43.5k93355
asked Nov 28 at 23:28
Laura1999
202
asked Nov 28 at 23:28
Laura1999
202
asked Nov 28 at 23:28
Laura1999
202
202
4
$4$ comes from the "..4-element subset .." and $8$ from $10$ elements ninus the $2$ excluded
– G Cab
Nov 28 at 23:31
Ho yeah sorry it's my mistake!
– Laura1999
Nov 28 at 23:39
Enumeration problems should be tagged combinatorics rather than probability.
– N. F. Taussig
Nov 28 at 23:47
add a comment |
4
$4$ comes from the "..4-element subset .." and $8$ from $10$ elements ninus the $2$ excluded
– G Cab
Nov 28 at 23:31
Ho yeah sorry it's my mistake!
– Laura1999
Nov 28 at 23:39
Enumeration problems should be tagged combinatorics rather than probability.
– N. F. Taussig
Nov 28 at 23:47
4
4
$4$ comes from the "..4-element subset .." and $8$ from $10$ elements ninus the $2$ excluded
– G Cab
Nov 28 at 23:31
$4$ comes from the "..4-element subset .." and $8$ from $10$ elements ninus the $2$ excluded
– G Cab
Nov 28 at 23:31
Ho yeah sorry it's my mistake!
– Laura1999
Nov 28 at 23:39
Ho yeah sorry it's my mistake!
– Laura1999
Nov 28 at 23:39
Enumeration problems should be tagged combinatorics rather than probability.
– N. F. Taussig
Nov 28 at 23:47
Enumeration problems should be tagged combinatorics rather than probability.
– N. F. Taussig
Nov 28 at 23:47
add a comment |
1 Answer
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Combination formula
$$C_{p}^{n} = frac{A_{p}^{n}}{p!} = frac{n!}{p!(n-p)!}$$
In your situation $p$ is the $4$ element subset, you replace each unknown in the general formula by its value and you get $70$ as an answer.
answered Nov 28 at 23:48
Blg Khalil
283
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Combination formula
$$C_{p}^{n} = frac{A_{p}^{n}}{p!} = frac{n!}{p!(n-p)!}$$
In your situation $p$ is the $4$ element subset, you replace each unknown in the general formula by its value and you get $70$ as an answer.
answered Nov 28 at 23:48
Blg Khalil
283
add a comment |
Combination formula
$$C_{p}^{n} = frac{A_{p}^{n}}{p!} = frac{n!}{p!(n-p)!}$$
In your situation $p$ is the $4$ element subset, you replace each unknown in the general formula by its value and you get $70$ as an answer.
answered Nov 28 at 23:48
Blg Khalil
283
add a comment |
Combination formula
$$C_{p}^{n} = frac{A_{p}^{n}}{p!} = frac{n!}{p!(n-p)!}$$
In your situation $p$ is the $4$ element subset, you replace each unknown in the general formula by its value and you get $70$ as an answer.
answered Nov 28 at 23:48
Blg Khalil
283
Combination formula
$$C_{p}^{n} = frac{A_{p}^{n}}{p!} = frac{n!}{p!(n-p)!}$$
In your situation $p$ is the $4$ element subset, you replace each unknown in the general formula by its value and you get $70$ as an answer.
answered Nov 28 at 23:48
Blg Khalil
283
answered Nov 28 at 23:48
Blg Khalil
283
answered Nov 28 at 23:48
Blg Khalil
283
answered Nov 28 at 23:48
Blg Khalil
283
283
add a comment |
add a comment |
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4
$4$ comes from the "..4-element subset .." and $8$ from $10$ elements ninus the $2$ excluded
– G Cab
Nov 28 at 23:31
Ho yeah sorry it's my mistake!
– Laura1999
Nov 28 at 23:39
Enumeration problems should be tagged combinatorics rather than probability.
– N. F. Taussig
Nov 28 at 23:47