Question

The total number of proper subsets for \[\left\{ {{\rm{a}},{\rm{b}},{\rm{c}},{\rm{d}}} \right\}\]are?
A.17
B.7
C.15
D.20

Answer 1

Hint: Look at definition of proper subsets for a set. Then apply the definition of this to the given set. Try to find the number of proper subsets possible. Use the concept of possibility to make the solution simple.

Complete step-by-step answer:
Proper subset: - A proper subset of a set A is a subset of A that is not equal to A. In other words, if B is a proper subset of A, then all elements of B are in A but A contains at least one element that is not B.
Given set in the question, can be written in the form:
\[\{ {\rm{A}},{\rm{B}},{\rm{C,D}})\]
The possibilities of A being in subset = 2 (Yes or No)
The possibilities of B being in subset = 2 (Yes or No)
The possibilities of C being in subset = 2 (Yes or No)
The possibilities of D being in subset = 2 (Yes or No)
By looking at both the definitions we can say that we have to use product rules. By using that rule, we can say that possibilities as;
\[Total{\rm { possibilities = }}{{\rm{2}}^{\rm{4}}}\]
As we know we have to remove the set itself and the empty set to get proper subsets. We get a number of proper subsets\[ = {2^4} - 2 = 14\].

Note: Don’t forget to subtract the 2 improper subsets. Alternate method is to list them one by one {A}, {B}, {C}, {D}, {AB}, {BC}, {CD}, {BD}, {AD}, {AC}, {ABC}, {ACD}, {BCD} {ABCD}. So, total 14 sets possible or another method is we have to select 1, 2, 3 elements from 4. So, \[total = {}^4{C_1} + {}^4{C_2} + {}^4{C_3} = 4 + 6 + 4 = 14\].