# Set Theory, Calendars, Clocks and Binomial Theorem

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## Set Theory: Number of Elements

Set P comprises all multiples of 4 less than 500. Set Q comprises all odd multiples of 7 less than 500, Set R comprises all multiples of 6 less than 500. How many elements are present in ${P\cup Q\cup R}$ ?
1. 202
2. 243
3. 228
4. 186

Choice A. 202

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## Detailed Solution

This question involves both Number Systems and Set Theory.

Set P = {4, 8, 12, ….496} ${\mapsto}$ 124 elements {all elements from 1 x 4 to 124 x 4}

Set Q = {7, 21, 35, 49,……497} ${\mapsto}$ {7 × 1, 7 × 3, 7 × 5, ….. 7 × 71} ${\mapsto}$ 36 elements.

Set R = {6, 12, 18, 24, …..498} ${\mapsto}$ {6 × 1, 6 × 2, 6 × 3, ….. 6 ×83} ${\mapsto}$ 83 elements.

Sets P and R have only even numbers; set Q has only odd numbers. So,

${P\cap Q}$ = Null set

${ Q\cap R}$ = Null set

${P\cap Q\cap R}$ = Null set

So, If we find ${P\cap R}$ , we can plug into the formula and get ${P\cup Q\cup R}$

${P\cap R}$ = Set of all multiples of 12 less than 500 = {12, 24, 36,…..492 = {12 × 1, 12 × 2 , 12 × 3, …12 ×41} ${\mapsto}$ This has 41 elements

${P\cup Q\cup R}$ = P + Q + R – (${P\cap Q}$) – (${ Q\cap R}$) – (${R\cap P}$) + (${P\cap Q\cap R}$)

${P\cup Q\cup R}$ = 124 + 36 + 83 – 0 – 0 – 41 + 0 = 202

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