quiz8ans - (500 isn't cubefree, so it doesn't matter if we...

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Quiz #8 for Math 55, Nov. 21 2002 1. Find the probability that when four numbers from 1 to 100, inclusive, are picked at random {em with no repetitions allowed}, either all are odd, all are divisible by 3, or all are divisible by 5. This is not exclusive-or; (9,15,21,33) should be counted even though all are odd and all are divisible by 3. A. Call these sets O, T, F. Then |O| = 50^4, |T| = 33^4, |F| = 20^4, |O intersect T| = 17^4, |O intersect F| = 10^4, |F intersect T| = 6^4, |O intersect F intersect T| = 3^4. So the answer is (|O| + |T| + |F| - |OT| - |OF| - |FT| + |OFT|) / 100^4. 2. How many cubefree integers are there less than 500? (You may be interested to know that 7^3 = 343.) A number is cubefree if it's not divsible by any n^3 for n more than 1.
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Unformatted text preview: (500 isn't cubefree, so it doesn't matter if we say less than 500 or less than or equal to 500, by the way.) A. To get the cubefree ones, we throw away the ones divisible by 2^3, 3^3, 5^3, and 7^3. Then we have to add one thing back: the one number less than 500 that's divisible by two of these, 2^3 and 3^3. So it's 499 - floor(499/2^3) - floor(499/3^3) - floor(499/5^3) - floor(499/7^3) + 1. (If you want, that's 499 - 62 - 55 - 3 - 1 + 1.) 3. How many ways can 0,1,2,3,4,5,6,7,8,9 be rearranged so that no even digit is in its final position? A. We basically follow the derangement derivation, and get 10! - (5 choose 1) 9! + (5 choose 2) 8! - (5 choose 3) 7! + (5 choose 4) 6! - (5 choose 5) 5!....
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