It's 5*9*10 because there are 900 three digit numbers between 100 and 999 and half of these are even, therefore 900/2 = 450.
Hi there I have a question...
Can anyone explain me this question which is along with its answer!
Count the set S of 3 digit numbers which begin or end
with an even digit.
Assume that 0 is even but a number cannot begin with 0.
The set is the union of the two subsets:
• The set B of three digit numbers that begin
with 2, 4, 6 or 8.
This set has cardinality
(4)(10)(10).
(why?)
• The set C of three digit numbers that end with
0, 2, 4, 6, or 8 and do not begin with 0.
This set has cardinality
(5)(9)(10). <<How come here it is (5)(9)(10) instead of (5)(10)(10)?
(why?)
• Now we use the inclusion-exclusion principle
to eliminate the overlap of sets B and C.
Their intersection:
The 3 digit numbers that begin with 2, 4, 6, or 8 and end
with 0, 2, 4, 6, or 8.
The intersection has the cardinality
(4)(10)(5)
Hence the cardinality is
400 + 450 - 200 = 650.
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Hello, robocop_911!
The first digit can be any of {2,4,6,8}: .4 choices.Count the set of 3-digit numbers which begin or end with an even digit.
Assume that 0 is even but a number cannot begin with 0.
The set is the union of the two subsets:
• The set of 3-digit numbers that begin with 2, 4, 6 or 8.
The second digit can be any of the 10 digits.
The third digit can be any of the 10 digits.
Hence, there are: . numbers in set
The first digit must not be zero: .9 choices.• The set of 3-digit numbers that end with {0,2,4,6,8} and do not begin with 0.
The second digit can be any digit: .10 choices.
The last digit is even: .5 choices.
Hence, there are: . numbers in set
The first digit is from {2,4,6,8}: .4 choices.• Now we use the inclusion-exclusion principle
to eliminate the overlap of sets B and C.
Their intersection:
The 3-digit numbers that begin with {2,4,6,8} and end with {0,2,4,6,8}.
The second digit can be any digit: .10 choices.
The last digit is from {0,2,4,6,8}: .5 choices.
Hence, there are: . numbers in the overlap .
Therefore: .