Question

A is a non-empty set having n elements. P and Q are two subsets of A, such that P is a subset of Q. Find the number of ways of choosing the subsets P and Q.

1. 4 ⁿ
2. 3 ⁿ
3. 2 ⁿ
4. n ²
 

Comments

Hey, what if they mention proper subset instead of subset, then will it be 2^n ?

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I think n*3^(n-1)

Reason : Atleast one element will be there that will go to set Q and not P. This one element can be chosen in n ways.

Remaining n-1 elements have three options to go to(Q and P , Q  but not P , not Q and not P).

This By product rule:

n*3^(n-1) ways are there.

Answer 1

i Think 3^n because every element of a has three choices .

either it should join p. then it should also come in q.

or join q then there is no restriction that it should belong to p also. fine.

or it should not appear anywhere. so n elements have a choice of 3 = 3^n

Answer 2

Lets take example let A={1,2}
Q can be 0 element set, then no of 0 element sets =1 .. Then its subsets possible ie P
=2^0 =1

Q can be  1 element set, no of such sets s =2 .. no of subsets in each = 2^1  

Q can be 2 element sets =1..
no of subsets possible =2^2

Total combinations=1+4+4=9

Since correct  option must be true for all cases including 2 element set , correct ans is 3 ^n

Answer 3

2.3^n