If $A = a $, $B = b, c $ and $C = d,e,f,g,h,i,j $, find (i) $P(A times B)$; (ii) $|P(B times C)|$.
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
If $A = a $, $B = b, c $ and $C = d,e,f,g,h,i,j $, find
(i) $P(A times B)$;
(ii) $|P(B times C)|$.
$A times B = (a, b), (a, c) $
$P(A times B) = emptyset, (a, b) , (a, c) , (a, b), (a, c) $
$|B times C| = |B| times |C| = 14$. So I think that $|P(B times C)| = 2^14$.
Is this correct? Thank you.
elementary-set-theory
add a comment |Â
up vote
0
down vote
favorite
If $A = a $, $B = b, c $ and $C = d,e,f,g,h,i,j $, find
(i) $P(A times B)$;
(ii) $|P(B times C)|$.
$A times B = (a, b), (a, c) $
$P(A times B) = emptyset, (a, b) , (a, c) , (a, b), (a, c) $
$|B times C| = |B| times |C| = 14$. So I think that $|P(B times C)| = 2^14$.
Is this correct? Thank you.
elementary-set-theory
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $A = a $, $B = b, c $ and $C = d,e,f,g,h,i,j $, find
(i) $P(A times B)$;
(ii) $|P(B times C)|$.
$A times B = (a, b), (a, c) $
$P(A times B) = emptyset, (a, b) , (a, c) , (a, b), (a, c) $
$|B times C| = |B| times |C| = 14$. So I think that $|P(B times C)| = 2^14$.
Is this correct? Thank you.
elementary-set-theory
If $A = a $, $B = b, c $ and $C = d,e,f,g,h,i,j $, find
(i) $P(A times B)$;
(ii) $|P(B times C)|$.
$A times B = (a, b), (a, c) $
$P(A times B) = emptyset, (a, b) , (a, c) , (a, b), (a, c) $
$|B times C| = |B| times |C| = 14$. So I think that $|P(B times C)| = 2^14$.
Is this correct? Thank you.
elementary-set-theory
asked Aug 26 at 13:59
Wyuw
1838
1838
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Yes, you have answered the question correctly.
In general, $P(X)=2^$ where $P(X)$ denote the power set of $X$ because for each possible subset we decide if it is in that particular subset.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes, you have answered the question correctly.
In general, $P(X)=2^$ where $P(X)$ denote the power set of $X$ because for each possible subset we decide if it is in that particular subset.
add a comment |Â
up vote
2
down vote
accepted
Yes, you have answered the question correctly.
In general, $P(X)=2^$ where $P(X)$ denote the power set of $X$ because for each possible subset we decide if it is in that particular subset.
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes, you have answered the question correctly.
In general, $P(X)=2^$ where $P(X)$ denote the power set of $X$ because for each possible subset we decide if it is in that particular subset.
Yes, you have answered the question correctly.
In general, $P(X)=2^$ where $P(X)$ denote the power set of $X$ because for each possible subset we decide if it is in that particular subset.
edited Aug 26 at 14:15
answered Aug 26 at 14:02
Siong Thye Goh
80.7k1453102
80.7k1453102
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2895065%2fif-a-a-b-b-c-and-c-d-e-f-g-h-i-j-find-i-pa%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password