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How many 9-digit codes can be formed if the code starts with 000 or ends with 1?
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To solve the problem, we should calculate separately
- the number of the 9-digit codes that start with 000;
- the number of the 9-digit codes that end with 1;
- and then calculate the number of the codes in the union of these two sub-sets.
(1) The number of the 9-digit codes that start with 000 is = ,
since the first three digits are just pre-determined, and only 6 remaining positions
each may have any of 10 digits.
(2) The number of the 9-digit codes that end with is = ,
since the last digit is just pre-determined, and only 8 remaining positions
each may have any of 10 digits.
(3) To complete the solution, apply the formula for the union
= + - .
equals to , since the intersection is the set of all codes
that have pre-determined 000 in the first 3 positions and 1 in the last position,
having all 5 = 9-3-1 positions free for any of 10 digits.
THEREFORE, the final formula and the final answer are
= + - = 1,000,000 + 100,000,000 - 100,000 = 100,900,000.
ANSWER. There are 100,900,000 such codes.
Solved.
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In short form and for your better understanding:
add the numbers of codes of each type; then subtract the number of codes
in the intersection, because otherwise you count them twice.