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in books i found that answer of this question is 3^5 because there are 3 ways for each element but one of the ways say that that the element not belong to both Y AND z so then how we can write a order pair of it so hoe\w we cam consider the the third case

in books i found that answer of this question is 3^5 because there are 3 ways for each element but one of the ways say that that the element not belong to both Y AND z so then how we can write a order pair of it so hoe\w we cam consider the the third case

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Grade:11

1 Answers

Arun
25758 Points
4 years ago
If y {} then z can have 25 subsets,  but one would be a null set therefore total 1*31

If y has one element then it could be in 5C1 ways, z can have 24 elements, that is 5*16 =80

If y has 2 elementsthen it could be in 5C2 ways and z can have 23 elements, that is 10*8=80

If y has 3 elements then it could be in 5C3 ways and z could have 22 elements, that is 10*4=40

If y has 4 elements then it could be in 5C4 ways and z could have 21 elements, that is 5*2=10

If y has 5 elements then it could be in 5C5 ways and z could have 21 elements, that is 1*2=2

Add all the cases.
You get as 
243 which is nothing but 3^5

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