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# 15b_01f_ms - Math 15B Midterm Exam Solutions 26 October...

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Math 15B Midterm Exam Solutions 26 October 2001 1. Give an example of each of the following or explain why it cannot be done. (a) A bijection from { 1 , 2 , 3 , 4 } to { a, b, c } . A. Impossible. A bijection has it’s range and domain the same size. (b) A permutation of { 1 , 2 , 3 , 4 , 5 , 6 , 7 } that has a cycle of length 4 and also has a cycle of length 5. A. Impossible. The sum of cycle lengths is the size of the set being permuted and 4 + 5 > 7. 2. A committee contains 7 women and 6 men. We want to form a subcommittee with 5 of these people. (a) How many ways can this be done? A. ( 7+6 5 ) (b) How many ways can this be done if the subcommittee must contain at least 2 women and at least 2 men? A. There are either (2 men AND 3 women) OR (3 men AND 2 women). By the Rules of Sum and Product, the answer is ( 6 2 ) × ( 7 3 ) + ( 6 3 ) × ( 7 2 ) . Choosing two of each sex and then a fifth committee member via ( 6 2 )( 7 2 )( 9 1 ) over- counts. For example, if there are 3 women on the committee, the committee is counted 3 times depending on which woman is selected as the fifth committee member.

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15b_01f_ms - Math 15B Midterm Exam Solutions 26 October...

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