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COT3100Counting03

COT3100Counting03 - Example Counting Questions(All taken...

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Example Counting Questions (All taken from past Foundation Exams) Martian Life Intelligent life has finally been discovered on Mars! Upon further study, linguistic researchers have determined several rules to which the Martian language adheres: 1) The Martian alphabet is composed of three symbols: a, b, and c. 2) Each word in the language is a concatenation of four of these symbols. 3) Each command in the language is composed of three words. a) How many distinct commands can be created if words in a single command can be repeated and two commands are identical only if the three words AND the order in which the words appear are identical? (Thus, the commands "aaca baaa aaca" and "baaa aaca aaca" are two DIFFERENT commands.) Total of 12 symbols in a command. For each of these symbols, we have 3 choices without any restrictions. These choices are independent of one another, so the total number of commands we have is 3 12 .

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b) How many distinct commands can be created if word position does not affect meaning and a given word may appear at most once in a single command? (Thus, "abca bbac abbb" and "bbac abbb abca" should NOT count as different commands, and "aaca baaa aaca" is an INVALID command.) Since we are not allowed to repeat words and word order doesn't matter, we are essentially choosing three words out of the a possible number of words. Thus, we must first figure out the possible number of words. There are three choices for each of four symbols, using the multiplication principle as we did in part a, we have 3 4 = 81 possible Martian words. Of these, we can choose three to make a valid command. Thus, the total number of possible commands here is 81 C 3 = (81)(80)(79)/6 = 85320 Counting Numbers Consider six-digit numbers with all distinct digits that do NOT start with 0. Answer the following questions about these numbers. Leave the answer in factorial form. a) How many such numbers are there? b) How many of these numbers contain a 3 but not 6? c) How many of these numbers contain either 3 or 6 (or both)? a) There are 9 choices for the first digit, and then 9 choices for the second digit (0 has been added as a choice), 8 for the third, 7 for the fourth, 6 for the fifth, and 5 for the sixth. Total = (9)(9) (8)(7)(6)(5) = 9(9!)/4! = 136080.
b) We need to separate the counting into two categories 1) 3 is the first digit 2) 3 is NOT the first digit For the first category, we have one choice for the first digit, followed by 8 choices for the second digit (not 3 or 6), 7 choices for the third digit, 6 choices for the fourth digit, 5 choices for the fifth digit and 4 choices for hte sixth digit. Total = (8)(7)(6)(5)(4) = 8!/3! For the second category, we have 7 choices for the first digit (not 0, 3, or 6), now we must guarantee that a 3 is picked. There are five PLACES to place the 3. For the remaining 4 digits, we have 7 choices, 6 choices, 5 choices and 4 choices, respectively for each of these. (To see this, imagine the 3 was placed 2nd.

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