Quiz #8 for Math 55, Nov. 21 2002
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1. Find the probability that when four numbers from 1 to 100, inclusive,
are picked at random {em with no repetitions allowed}, either 
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<li> all are odd, 
<li> all are divisible by 3, or 
<li> all are divisible by 5.
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This is not exclusive-or; (9,15,21,33) should be counted even though all are
odd <i>and</i> all are divisible by 3.

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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.
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So the answer is (|O| + |T| + |F| - |OT| - |OF| - |FT| + |OFT|) / 100^4.

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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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(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.)
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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.
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So it's 499 - floor(499/2^3) - floor(499/3^3) - floor(499/5^3) - floor(499/7^3)
+ 1.
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(If you want, that's 499 - 62 - 55 - 3 - 1 + 1.)


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3. How many ways can 0,1,2,3,4,5,6,7,8,9 be rearranged so that 
no <i>even</i> digit is in its final position?

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A. We basically follow the derangement derivation, and get
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10! - (5 choose 1) 9! + (5 choose 2) 8! 
    - (5 choose 3) 7! + (5 choose 4) 6! - (5 choose 5) 5!.