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QJEEM23Ma1fa1: JEE Main 2023 Mathematics - Class 11 JEEM23Ma1fa1 Solution: The number of 3-digit numbers, that are divisible by either 2…

The number of 3-digit numbers, that are divisible by either 2 or 3 but not divisible by 7, is ____________.

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JEE Main 2023 (Online) 1st February Morning Shift Mathematics - Permutations and Combinations Question 75 English Explanation

A=A= Numbers divisible by 2

B=B= Numbers divisible by 3

C=C= Numbers divisible by 7

n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B)=n(A)+n(B)-n(A \cap B)

=n(2)+n(3)−n(6)=n(2)+n(3)-n(6)

n(A)=n(2)=100,102….,998=450n(A)=n(2)=100,102 \ldots ., 998=450

n(B)=n(3)=102,105,….,999=30n(B)=n(3)=102,105, \ldots ., 999=30

n(A∩B)=n(6)=102,108,….,996=150n(A \cap B)=n(6)=102,108, \ldots ., 996=150

n(2n(2 or 3 )=450+300−150=600)=450+300-150=600

Now,

n( A∩C)=n(14)=112,126,….,994=64n(\mathrm{~A} \cap C)=n(14)=112,126, \ldots ., 994=64

n(A∩B∩C)=n(42)=126,168,….,966=21n(A \cap B \cap C)=n(42)=126,168, \ldots ., 966=21

n(B∩C)=n(21)=105,126,……,987,=43n(B \cap C)=n(21)=105,126, \ldots \ldots, 987,=43

n(2n(2 or 3 not by 7 )=600−[64+43−21])=600-[64+43-21]

=514=514

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