How many subsets of size 4 are there?
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All I can think of is there are $2^8$ subsets of the set. Then, there are $2^5$ subsets of size $5.$ So $2^8-2^5$ gives you the number of subsets of four elements. How do I even approach this problem?
combinatorics discrete-mathematics
edited Dec 14 '18 at 17:07
user376343
3,7783827
asked Dec 14 '18 at 17:05
PumpkinpeachPumpkinpeach
628
$endgroup$
add a comment |
$begingroup$
All I can think of is there are $2^8$ subsets of the set. Then, there are $2^5$ subsets of size $5.$ So $2^8-2^5$ gives you the number of subsets of four elements. How do I even approach this problem?
combinatorics discrete-mathematics
edited Dec 14 '18 at 17:07
user376343
3,7783827
asked Dec 14 '18 at 17:05
PumpkinpeachPumpkinpeach
628
$endgroup$
4
$begingroup$
There aren't $2^5$ subsets of size $5$...there are $binom 85$. And why would the number of subsets of size $5$ plus the number of size $4$ equal the total number of subsets?
$endgroup$
– lulu
Dec 14 '18 at 17:06
add a comment |
$begingroup$
All I can think of is there are $2^8$ subsets of the set. Then, there are $2^5$ subsets of size $5.$ So $2^8-2^5$ gives you the number of subsets of four elements. How do I even approach this problem?
combinatorics discrete-mathematics
edited Dec 14 '18 at 17:07
user376343
3,7783827
asked Dec 14 '18 at 17:05
PumpkinpeachPumpkinpeach
628
$endgroup$
All I can think of is there are $2^8$ subsets of the set. Then, there are $2^5$ subsets of size $5.$ So $2^8-2^5$ gives you the number of subsets of four elements. How do I even approach this problem?
combinatorics discrete-mathematics
combinatorics discrete-mathematics
edited Dec 14 '18 at 17:07
user376343
3,7783827
asked Dec 14 '18 at 17:05
PumpkinpeachPumpkinpeach
628
edited Dec 14 '18 at 17:07
user376343
3,7783827
asked Dec 14 '18 at 17:05
PumpkinpeachPumpkinpeach
628
edited Dec 14 '18 at 17:07
user376343
3,7783827
edited Dec 14 '18 at 17:07
user376343
3,7783827
edited Dec 14 '18 at 17:07
user376343
3,7783827
3,7783827
asked Dec 14 '18 at 17:05
PumpkinpeachPumpkinpeach
628
asked Dec 14 '18 at 17:05
PumpkinpeachPumpkinpeach
628
asked Dec 14 '18 at 17:05
PumpkinpeachPumpkinpeach
628
628
4
$begingroup$
There aren't $2^5$ subsets of size $5$...there are $binom 85$. And why would the number of subsets of size $5$ plus the number of size $4$ equal the total number of subsets?
$endgroup$
– lulu
Dec 14 '18 at 17:06
add a comment |
4
$begingroup$
There aren't $2^5$ subsets of size $5$...there are $binom 85$. And why would the number of subsets of size $5$ plus the number of size $4$ equal the total number of subsets?
$endgroup$
– lulu
Dec 14 '18 at 17:06
4
4
$begingroup$
There aren't $2^5$ subsets of size $5$...there are $binom 85$. And why would the number of subsets of size $5$ plus the number of size $4$ equal the total number of subsets?
$endgroup$
– lulu
Dec 14 '18 at 17:06
$begingroup$
There aren't $2^5$ subsets of size $5$...there are $binom 85$. And why would the number of subsets of size $5$ plus the number of size $4$ equal the total number of subsets?
$endgroup$
– lulu
Dec 14 '18 at 17:06
add a comment |
1 Answer
1
active
oldest
votes
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Hint: Make a grid where the rows are labeled one to eight, and the columns are labeled with the subsets. Place a check mark in a box if the subset at that column contains the number at that row. Then every row has three check marks, and every column has four...
answered Dec 14 '18 at 17:12
Mike EarnestMike Earnest
22.6k12051
$endgroup$
$begingroup$
Do you mention $70$ columns ...
$endgroup$
– user376343
Dec 14 '18 at 17:21
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No, there is only a column for each subset being considered. @user376343
$endgroup$
– Mike Earnest
Dec 14 '18 at 17:22
$begingroup$
I was thinking of 4-element subsets, total number. Because at the beginning we don't know the convenient collection of subsets.
$endgroup$
– user376343
Dec 14 '18 at 17:26
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
Hint: Make a grid where the rows are labeled one to eight, and the columns are labeled with the subsets. Place a check mark in a box if the subset at that column contains the number at that row. Then every row has three check marks, and every column has four...
answered Dec 14 '18 at 17:12
Mike EarnestMike Earnest
22.6k12051
$endgroup$
$begingroup$
Do you mention $70$ columns ...
$endgroup$
– user376343
Dec 14 '18 at 17:21
$begingroup$
No, there is only a column for each subset being considered. @user376343
$endgroup$
– Mike Earnest
Dec 14 '18 at 17:22
$begingroup$
I was thinking of 4-element subsets, total number. Because at the beginning we don't know the convenient collection of subsets.
$endgroup$
– user376343
Dec 14 '18 at 17:26
add a comment |
$begingroup$
Hint: Make a grid where the rows are labeled one to eight, and the columns are labeled with the subsets. Place a check mark in a box if the subset at that column contains the number at that row. Then every row has three check marks, and every column has four...
answered Dec 14 '18 at 17:12
Mike EarnestMike Earnest
22.6k12051
$endgroup$
$begingroup$
Do you mention $70$ columns ...
$endgroup$
– user376343
Dec 14 '18 at 17:21
$begingroup$
No, there is only a column for each subset being considered. @user376343
$endgroup$
– Mike Earnest
Dec 14 '18 at 17:22
$begingroup$
I was thinking of 4-element subsets, total number. Because at the beginning we don't know the convenient collection of subsets.
$endgroup$
– user376343
Dec 14 '18 at 17:26
add a comment |
$begingroup$
Hint: Make a grid where the rows are labeled one to eight, and the columns are labeled with the subsets. Place a check mark in a box if the subset at that column contains the number at that row. Then every row has three check marks, and every column has four...
answered Dec 14 '18 at 17:12
Mike EarnestMike Earnest
22.6k12051
$endgroup$
Hint: Make a grid where the rows are labeled one to eight, and the columns are labeled with the subsets. Place a check mark in a box if the subset at that column contains the number at that row. Then every row has three check marks, and every column has four...
answered Dec 14 '18 at 17:12
Mike EarnestMike Earnest
22.6k12051
answered Dec 14 '18 at 17:12
Mike EarnestMike Earnest
22.6k12051
answered Dec 14 '18 at 17:12
Mike EarnestMike Earnest
22.6k12051
answered Dec 14 '18 at 17:12
Mike EarnestMike Earnest
22.6k12051
22.6k12051
$begingroup$
Do you mention $70$ columns ...
$endgroup$
– user376343
Dec 14 '18 at 17:21
$begingroup$
No, there is only a column for each subset being considered. @user376343
$endgroup$
– Mike Earnest
Dec 14 '18 at 17:22
$begingroup$
I was thinking of 4-element subsets, total number. Because at the beginning we don't know the convenient collection of subsets.
$endgroup$
– user376343
Dec 14 '18 at 17:26
add a comment |
$begingroup$
Do you mention $70$ columns ...
$endgroup$
– user376343
Dec 14 '18 at 17:21
$begingroup$
No, there is only a column for each subset being considered. @user376343
$endgroup$
– Mike Earnest
Dec 14 '18 at 17:22
$begingroup$
I was thinking of 4-element subsets, total number. Because at the beginning we don't know the convenient collection of subsets.
$endgroup$
– user376343
Dec 14 '18 at 17:26
$begingroup$
Do you mention $70$ columns ...
$endgroup$
– user376343
Dec 14 '18 at 17:21
$begingroup$
Do you mention $70$ columns ...
$endgroup$
– user376343
Dec 14 '18 at 17:21
$begingroup$
No, there is only a column for each subset being considered. @user376343
$endgroup$
– Mike Earnest
Dec 14 '18 at 17:22
$begingroup$
No, there is only a column for each subset being considered. @user376343
$endgroup$
– Mike Earnest
Dec 14 '18 at 17:22
$begingroup$
I was thinking of 4-element subsets, total number. Because at the beginning we don't know the convenient collection of subsets.
$endgroup$
– user376343
Dec 14 '18 at 17:26
$begingroup$
I was thinking of 4-element subsets, total number. Because at the beginning we don't know the convenient collection of subsets.
$endgroup$
– user376343
Dec 14 '18 at 17:26
add a comment |
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4
$begingroup$
There aren't $2^5$ subsets of size $5$...there are $binom 85$. And why would the number of subsets of size $5$ plus the number of size $4$ equal the total number of subsets?
$endgroup$
– lulu
Dec 14 '18 at 17:06