solutions 6

Discrete Mathematics and Its Applications with MathZone

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Unformatted text preview: ICS 6D ( Discrete Mathematic): Solution to Homework #6 Section 5.1 4. 12 * 2 * 3 8. Heres one way to count: Take the number of all 3-letter initials, 26 3 , and substruct the number of 3-letter initials that have one repetition, which is 3 * 26 * 25 (because theres three possibilities for where the non-doubled letter is, 26 choices for the doubled letter, and 25 choices for the non-doubled letter), and 3-letter initials, which is 26. So the answer is 26 3- (3 * 26 * 25 + 26). 20. (a) b 999 / 7 c . 1 (b) b 999 / 7 c-b 999 / 77 c , because you have to substruct numbers that are divisible by both 7 and 11, which is the same set as the set of numbers that are divisble by 77. (c) b 999 / 77 c , see above. (d) b 999 / 7 c + b 999 / 11 c - b 999 / 77 c , using the inclusion-exclusion principle and using part (c). (e) ( b 999 / 7 c - b 999 / 77 c ) + ( b 999 / 11 c - b 999 / 77 c ) = b 999 / 7 c + b 999 / 11 c - 2 b 999 / 77 c . Its the same argument as in part (b), but used twice, first for all numbers divisible by 7 and not 11, and then for all those that are divisible by 11 but not by 7. (f) 999- ( b 999 / 7 c + b 999 / 11 c - b 999 / 77 c ), by part (d). (g) 9 + 9 * 9 + 9 * 9 * 8, summing the counts of numbers with no repetitions among digits which have, respectively, exactly one digit, exactly two digits, and exactly three digits. Note that in each case the first digit has to be non-zero, so there is 9 choices, and the other digits can be any digit among all 10 digits except of the previously taken digits, so there are 9 choices of the second digit, 8 choices of the third digit, etc. (h) This one was tricky! As I went over it in the class, theres many incorrect paths one can fall into, but heres a way that works: Let o i be the count of i-digit numbers that satisfy this constraint, i.e. they are even and all their digits are different. The answer is o 1 + o 2 + o 3 . We have that o 1 = |{ 2 , 4 , 6 , 8 } = 4. We also have o 2 = 4 * 8 + 1 * 9, because if the number has two digits then if its last digit is non-zero (4 choices) then its first digit has 8 choices (any digit except zero and the one thats equal to the last digit), and if its last digit is zero (1 choice) then its first digit has 9 choices (any digit except zero). Finally we have o 3 = 4 * 8 * 8 + 1 * 9 * 8 because if the number has three digits then if its last digit is non-zero (4 choices) then its first digit has 8 choices (any digit except zero and the one thats chosen as the last digit) and its...
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solutions 6 - ICS 6D ( Discrete Mathematic): Solution to...

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