TIFR CSE 2012 | Part A | Question: 8

Question

How many pairs of sets $(A, B)$ are there that satisfy the condition $A, B \subseteq \left\{1, 2,...,5\right\}, A \cap B = \{\}?$

• A. $125$
• B. $127$
• C. $130$
• D. $243$
• E. $257$

Comments

Similar questions: GATE CSE 2003 | Question: 5, TIFR2010-A-18, GATE CSE 2021 Set 2 | Question: 50

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which means nco*2^0 + nc1*2^1 series is same as nco*2^5 + nc1*2^4

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let us take n =2 {a,b}

suppose there are two subsets A and B such that either A intersection B = phi

now,  a has three choices either go to a set A or go to set B or go no where, hence 3^1

also

b has three choices either go to a set A or go to set B or go no where, hence 3^1

 

so for 2 elements 3^1 * 3^1 = 3^2

thus , for n elements 3^n

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Detailed Video Explanation with Complete Analysis: https://www.youtube.com/watch?v=Mxxzmd-BQ18&t=1807s

Same Question in ISI two times:

• https://gateoverflow.in/321869/isi2015-mma-8  
• https://gateoverflow.in/432358/isi2024-mcs-pca-question-12  

Answer 1

Correct Option: $C$ $243$.

First take $A$ as $\emptyset$ and $B$ as power set of $\{1,2,3,4,5\} $ which is $2^5$. Then take $A$ as set of one element ex: when $A$=$\left \{ 1 \right \}$ then set $B$ could be any of the $2^4$ elements of power set. This will give us $16 \times 5$ . In similar fashion when $A$ consists of $2$ elements we get total pairs $5C2\times 2^3$; when $A$ is of three elements we get $40$; for $4$ elements we get $10$ pairs and when $A$ is of $5$ elements we get one pair which is $A$=$\left \{ 1,2,3,4,5 \right \}$ and $B=\left \{ \right \}.$ So, in total $=32+80+80+40+10+1=243$

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Alternative Solution:

for each element in $[n]$, you have 3 choices:

1. Include it in $A$ but not in $B$
2. Include it in $B$, but not in $A  $
3. Include it in neither

so this gives $3^n$ pairs.

 for set $\{1,2,,...,5\}$, $n=5$, $3^5=243$

Comments

^ Related concept used in https://gateoverflow.in/18496/tifr2010-a-18

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I have a doubt here.

 When we take A as set size =1, then we are taking set size B as 1/2/3/4/5

Example for A={1} we have B={2,3)(B set size=2) such that A intersection B = null

 

Now when A is having set size=2 .Then again we are taking A={(2,3)} and B={1}.

Aren't we counting it multiple times? Please clarify.

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A and B are two different sets.

$A = \lbrace 1,2 \rbrace \text{ and } B = \lbrace 3,4,5 \rbrace \\$
$A = \lbrace 3,4,5 \rbrace \text{ and } B = \lbrace 1,2 \rbrace \\$

are two different cases. 

Answer 2

Option (D) 243 ,is the correct answer.