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Unformatted text preview: WEEK 7 HOMEWORK SOLUTIONS 5.510 : a: ( 12+6 1 12 ) = 6188 b: ( 36+6 1 36 ) = 749398 c: This is the same as choosing 1 dozen croissants beyond the ones that are forced. Answer: 6188 d: 0 Broccolis: ( 24+5 1 24 ) = 20475, 1 broccoli: ( 23+5 1 23 ) = 17550, 2 broccolis: ( 22+5 1 22 ) = 14950. Total: 52975. e: ( 16+6 1 16 ) = 20349 f: With any number of broccolis: ( 15+6 1 15 ) = 15504, with at least 4 broccolis: ( 11+6 1 11 ) = 4368. With at most 3 broccolis: 15504 4368 = 11136. 5.515 : a: The number is the same as for solving x 1 + x 2 + x 3 + x 4 + x 5 = 20 with positive integer solutions (subtract one from x 1 for each solution). Answer: ( 24 20 ) = 10626. b: ( 15 11 ) = 1365. c: With x 1 ≥ 0 : ( 25 21 ) = 12650, with x 1 ≥ 11 : ( 14 10 ) = 1001. Answer: 12650 1001 = 11649. d: This is the same as the number of solutions to x 1 + x 2 + x 3 + x 4 + x 5 = 5, with 0 ≤ x 1 ≤ 3 , ≤ x 2 ≤ 2. Answer: ( 9 5 ) 20 = 106....
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This note was uploaded on 04/05/2010 for the course MATH 55 taught by Professor Strain during the Spring '08 term at Berkeley.
 Spring '08
 STRAIN
 Math

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