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Unformatted text preview: k1 committee members in ± n1 k1 ² . Combining the two we have n ± n1 k1 ² , possible choices. Part (d): Since all expressions count the same thing they must be equal and we have k ± n k ² = ( nk + 1) ± n k1 ² = n ± n1 k1 ² . Part (e): We have k ± n k ² = k n ! ( nk )! k ! = n ! ( nk )!( k1)! = n !( nk + 1) ( nk + 1)!( k1)! = ( nk + 1) ± n k1 ²...
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This note was uploaded on 02/25/2011 for the course STAT 418 taught by Professor G.jogeshbabu during the Winter '08 term at Pennsylvania State University, University Park.
 Winter '08
 G.JOGESHBABU
 Probability

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