You visited us 1 times! Enjoying our articles? Unlock Full Access!

Cardinality of Power Set of a Set

For a set K, ...

Question

For a set $K, if n (K) = m;$ then the number of proper subsets of $K =$

2^{m} - 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2^{m - 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2^{m}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $2^{m} - 1$
Given a set $K$ such that $n (K) = m$
Now improper set of any set is that set itself.
Since $2^{m}$ is the total number of subsets of any sets.
$∴$ Total number of proper subsets of a set $= 2^{m} - 1.$

Suggest Corrections

Similar questions

K, if n (K) = m;

K =

m

A = {a_{1}, a_{2}, . . ., a_{n}}

k

m

a_{4}

n

A

m

A

S = {1, 2, 3, . . . . ., 9}

k = 1, 2, . . . . ., 5

N_{k}

S

k

N_{1} + N_{2} + N_{3} + N_{4} + N_{5} =

S = {1, 2, 3, \dots, 9}

k = 1, 2, \dots, 5

N_{k}

S

k

N_{1} + N_{2} + N_{3} + N_{4} + N_{5} =

Join BYJU'S Learning Program