Class 11 Maths Sets Number of Elements in a set

Number of Elements in a set

If A, B and C are finite sets, then

  • n ( A ∪ B ) = n ( A ) + n ( B ) – n ( A ∩ B )

Explanation for n ( A B ) = n ( A ) + n ( B ) – n ( A ∩ B ):  Since the common elements A ∩ B is counted twice with both n ( A ) & n ( B )   , we subtract it.

  • n ( A ∪ B ∪ C ) = n ( A ) + n ( B ) + n ( C ) – n ( A ∩ B ) – n ( B ∩ C) – n ( A ∩ C ) + n ( A ∩ B ∩ C )

Refer ExamFear Video Lessons for explanation.

 

Numerical: In a school there are 20 teachers who teach mathematics or physics. Of these, 12 teach mathematics and 4 teach both physics and mathematics. How many teach physics ?

Solution: Let P denote Physics teachers & M denote Maths Teacher.

n(M) = 12

n(M ∪ P) = 20

n(M ∩ P) = 4

 Applying formula     n ( M ∪ P ) = n ( M ) + n (P ) – n ( M ∩ P )

Or 20 = 12 + n(P) – 4

Or n(P) = 12

We can also solve this with Venn diagram

 

 

 

Numerical: A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58 men and only three men got medals in all the three sports, how many received medals in exactly two of the three sports ?

We can represent the data using Venn diagram.

Steps

  • 3 men got all 3 medals, so n ( A ∩ B ∩ C ) will be 3. Thus put 3 in the region n ( A ∩ B ∩ C )
  • Lets count of orange, blue & pink region be a, b & c. These people received exactly 2 medals.
  • The Purple region will be 35-a-b, since 38 people got medals in football. Total count for football circle is 38. Similarly grey region will be 12-a-c & Green will be 17-b-c
  • Now it is total that total 58 men received these medals. That is n ( A ∪ B ∪ C ) = 58

Or 58 =(35-a-b) + a + (12-a-c) + 3 + b + (17-b-c) + c

Or a+b+c =9

Thus 9 people received 2 medals.

Thus we can say that 3 men received 3 medals, 9 men received 2 medals & 26 men received 1 medal.

 

For good explanation of these concepts, refer ExamFear video lessons.

 

 

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