Question

Let a set S={ 1 , 2 , 3 , 4 } . Now in how many ways subsets A , B and C can be selected from S such that
$A \cup B \cup C = S$
​

Answer 1

Answer:

51

Step-by-step explanation:

The given set, S = { 1 , 2 , 3 , 4 }

Here the Power set of S is given as :

$:\longmapsto$P(S) = { {1} , {2} , {3} , {4} , {1,2} , {1,3} , {1,4} , {2,3} , {2,4} , {3,4} , {1,2,3} , {1,2,4} , {1,3,4} , {2,3,4} , {1,2,3,4} , ф }

Now , we can select the sets A, B, C , such that $A \cup B \cup C = S$

The possible values are,

$\twoheadrightarrow$ A = {1,2,3,4} , B = ф , C = ф

$\twoheadrightarrow$ A = {1,2} , B = {3} , C = {4}

$\twoheadrightarrow$ A = {1,2} , B = {4} , C = {3}

$\twoheadrightarrow$ A = {2,3} , B = {1} , C = {4}

$\twoheadrightarrow$ A = {2,3} , B = {4} , C = {1}

$\twoheadrightarrow$ A = {3,4} , B = {1} , C = {2}

$\twoheadrightarrow$ A = {3.4} , B = {2} , C = {1}

$\twoheadrightarrow$ A = {1,4} , B = {2} , C = {3}

$\twoheadrightarrow$ A = {1,4} , B = {3} , C = {2}

$\twoheadrightarrow$ A = {1,2,3} , B = {4} , C = ф

$\twoheadrightarrow$ A = {1,2,3} , B = ф , C = {4}

$\twoheadrightarrow$ A = {1,2,4} , B = {3} , C = ф

$\twoheadrightarrow$ A = {1,2,4} , B = ф , C = {3}

$\twoheadrightarrow$ A = {1,3,4} , B = {2} , C = ф

$\twoheadrightarrow$ A = {1,3,4} , B = ф , C = {2}

$\twoheadrightarrow$ A = {2,3,4} , B = {1} , C = ф

$\twoheadrightarrow$ A = {2,3,4} , B = ф , C = {1}

These all in total are for A and are 17

For B, C these are also 17

Thus , total ways of selecting A, B, C such that $A \cup B \cup C = S$ is = 17 × 3 = 51

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Answer 2

Answer:

${ \boxed{\mathbb{\red{⛄answer♡࿐⛄}}}} \\ 51 \\ \\ </p><p></p><p> \underline{ \boxed{\mathbb{\red{⛄࿐⛄Step-by-step explanation:}}}} \\ \\ </p><p></p><p>The given set, S = { 1 , 2 , 3 , 4 } \\ </p><p></p><p>Here the Power set of S is given as : \\ </p><p></p><p>:\longmapsto:⟼ P(S) = { {1} , {2} , {3} , {4} , {1,2} , {1,3} , {1,4} , {2,3} , {2,4} , {3,4} , {1,2,3} , {1,2,4} , {1,3,4} , {2,3,4} , {1,2,3,4} , ф } \\ </p><p></p><p>Now , we can select the sets A, B, C , such that A \cup B \cup C = SA∪B∪C=S \\ </p><p></p><p>The possible values are, \\ </p><p></p><p>\twoheadrightarrow↠ A = {1,2,3,4} , B = ф , C = ф \\ </p><p></p><p>\twoheadrightarrow↠ A = {1,2} , B = {3} , C = {4} \\ </p><p></p><p>\twoheadrightarrow↠ A = {1,2} , B = {4} , C = {3} \\ </p><p></p><p>\twoheadrightarrow↠ A = {2,3} , B = {1} , C = {4} \\ </p><p></p><p>\twoheadrightarrow↠ A = {2,3} , B = {4} , C = {1}</p><p></p><p>\twoheadrightarrow↠ A = {3,4} , B = {1} , C = {2} \\ </p><p></p><p>\twoheadrightarrow↠ A = {3.4} , B = {2} , C = {1} \\ </p><p></p><p>\twoheadrightarrow↠ A = {1,4} , B = {2} , C = {3} \\ </p><p></p><p>\twoheadrightarrow↠ A = {1,4} , B = {3} , C = {2} \\ </p><p></p><p>\twoheadrightarrow↠ A = {1,2,3} , B = {4} , C = ф \\ </p><p></p><p>\twoheadrightarrow↠ A = {1,2,3} , B = ф , C = {4} \\ </p><p></p><p>\twoheadrightarrow↠ A = {1,2,4} , B = {3} , C = ф \\ </p><p></p><p>\twoheadrightarrow↠ A = {1,2,4} , B = ф , C = {3} \\ </p><p></p><p>\twoheadrightarrow↠ A = {1,3,4} , B = {2} , C = ф \\ </p><p></p><p>\twoheadrightarrow↠ A = {1,3,4} , B = ф , C = {2} \\ </p><p></p><p>\twoheadrightarrow↠ A = {2,3,4} , B = {1} , C = ф \\ </p><p></p><p>\twoheadrightarrow↠ A = {2,3,4} , B = ф , C = {1} \\ </p><p></p><p>These all in total are for A and are 17 \\ </p><p></p><p>For B, C these are also 17 \\ </p><p></p><p>Thus , total ways of selecting A, B, C such that A \cup B \cup C = SA∪B∪C=S is = 17 × 3 = 51 \\ </p><p></p><p>$

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