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COT3100C01, Fall 2002
Oct. 15, 2002,
correction on
S. Lang
Solution Keys to Assignment #3 (40 pts.)
3(c) in red added 10/18,
a
simpler solution to 2(b) in blue
added 10/28
1.
(12 pts.) Count the number of 7digit integers under each of the following restrictions:
(a) All digits are distinct.
(Thus, integers such as 1234567 and 3456980 are counted, but integers
such as 2345 and 2233445 are not counted.)
Answer:
9 * P(9, 6) = 9*9*8*7*6*5*4, where the first factor 9 counts the choices of a nonzero
digit (1 through 9) for the leftmost digit of the 7digit integer; the factor P(9, 6) counts the
number of permutations for the remaining 6 digits of the 7digit integer.
(b) The digits may be repeated (such as 2233456) but the integer must be divisible by 5.
(Hint: what
would be the last, rightmost digit for an integer to be divisible by 5?)
Answer:
9 * 10
5
* 2, where the first factor 9 counts the 9 choices (1 through 9) for the leftmost
digit; each of the middle 5 digits can be any of the 10 choices (0 through 9); finally, the
rightmost digit must be either 0 or 5, for two choices, for the integer to be divisible by 5.
(c) The digits may be repeated but digit 3 must be used (as part of the integer) at least twice.
(Hint:
consider the cases: digit 3 used twice, three times, etc., until 7 times, then count each case or
group separately.
Alternatively, consider the complementary question, i.e, if digit 3 is not used
or used exactly once, and count them separately.)
Answer:
We consider the complementary question.
First, we count those 7digit integers that
don’t use digit 3.
This count is 8 * 9
6
because the first digit can have 8 choices (1, 2, 4 through
9), each of the remaining 6 digits has 9 choices (0 though 9 excluding 3).
We then count those 7digit integers that contain digit 3 exactly once.
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