Mathematics - Online Test

First we will find the  number of three digit numbers (i.e, numbers from 100 to 999)which can be formed using the digits 0,1,2,3,4,5,6,7,8and 9 with repitition allowed .

Now we have the first place can be filled by any of the 9 digits other than 0 and since repetition is allowed the second and third can be filled by any of the ten digits.

Hence the total number of three digit numbers will be = 9×10×10=900

Now we will consider the case that the number does not have the digit 5.

Now the  first place can be filled by any of the 8 digits other than 0 and 5 and since repetition is allowed the second and third can be filled by any of the 9 digits other than 5.

Hence the total number of ways we can form a  three digit number with out 5 will be =8×9×9=648

Therefore the number of three digit numbers with atleast one 5=900−648=252

n(A) = 3

n(B)=6

n(AUB) = n(A)+n(B)-n(A∩B)

for minimum number of elements in A U B

n(A∩B)= 3

so n(AUB = 6