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222m1s-practice - Solution Ideas for 201001 Math 222...

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Solution Ideas for 201001 Math 222 Practice Midterm 1 Questions 1. (a) If the first digit is 7, there are ( 4 2 ) 9 2 such integers, and if the first digit isn’t 7 there are ( 4 3 ) 9 · 8 of them. Hence, by the AP, the total number is ( 4 2 ) 9 2 + ( 4 3 ) 9 · 8. (b) If the first digit is even, then the number of such integers is 4( ( 4 2 ) 5 4 + ( 4 3 ) 5 4 + ( 4 4 ) 5 4 ), and if the first digit is odd, there are 5( ( 4 3 ) 5 4 + ( 4 4 ) 5 4 ) of them. Hence, by the AP, the total number is 4 · 5 4 ( ( 4 2 ) + ( 4 3 ) + ( 4 4 ) ) + 5 5 ( ( 4 3 ) + ( 4 4 ) ). (c) 9 · 9 · 8 · 7 · 6 2. There are 61 choose 2 choices for the pair ( x 2 , x 3 ) – once the two numbers are known the one encountered first when travelling clockwise from 0 must be x 2 . There are then 60 choices for x 1 6 = x 2 . Hence the number of valid combinations is 60 ( 61 2 ) 3. 2 15 - 2 7 . 4. (a) The integer a ∈ { 1 , 2 , . . . , 10 } belongs to the subset S f if and only if f ( a ) = 1. (b) If f and g are different functions, then f ( a ) 6 = g ( a ) for some a { 1 , 2 , . . . , 10 } . Therefore a belongs to one of S f and S g (see (a) for notation), but does not belong to the other one. (c) Given S ⊆ { 1 , 2 , . . . , 10 } , define f by setting f ( a ) = 1 when a S and f ( a ) = 0 otherwise. 5. (a) 12! / 6. (b) Seat Gary on a fixed side of the table (2 ways) and seat everybody else relative to him. There are 5! ways to assign the remaining men to the remaining 5 sides of the table, and each of these men has 2 choices of seat. This leaves one seat on each side of the table to be
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