hw7sol - Math 55: Discrete Mathematics UC Berkeley, Spring...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 55: Discrete Mathematics UC Berkeley, Spring 2009 Solutions to Homework # 7 (due March 16, 2009) 5.5, #10: a) There are 6 types of croissants. This is a case of 12-combinations with repetition, or, equivalently, putting 12 indistinguishable objects in 6 distinguishable boxes. The number of possibilitites is ( 12+6- 1 12 ) = 6188. b) Similar to above, the number of possibilities is ( 36+6- 1 36 ) = 749398. c) We can suppose that the first dozen consist of 2 of each of the 6 kinds. Then the number of possibilities is the same as the number of ways of choosing the second dozen, which is 6188, the same as in part a. d) Choosing 24 croissants with exactly k of them broccoli is the same as choosing 24- k croissants among 5 kinds. Thus, the answer is ( 24+5- 1 24 ) + ( 23+5- 1 23 ) + ( 22+5- 1 22 ) = 52975. e) This is the same as choosing 24- 3- 5 = 16 croissants. The number of ways is ( 16+6- 1 16 ) = 20349. f) This is the same as choosing 24- 1- 2- 3- 1- 2 = 15 croissants with no more than 3 of them broccoli. Similar to part d, this is ( 15+5- 1 15 ) + ( 14+5- 1 14 ) + ( 13+5- 1 13 ) + ( 12+5- 1 12 ) = 11136. 5.5, #15 : a) The same as the non-negative integers wich add to 20: ( 20+5- 1 20 ) = 10626. b) The same as the non-negative integers adding to 11, of which there are ( 11+5- 1 11 ) = 1365. c) Similar to part a, the number of non-negative integers with x 1 11 is ( 10+5- 1 10 ) = 1001. The number of solutions with all non-negative in- tegers is ( 21+5- 1 21 ) = 12650. Subtracting, we get 11649. d) Similar to parts a and b, this is the same as the number of nonnega- tive integer solutions to x 1 + x 2 + x 3 + x 4 + x 5 = 5 subject to x 1 3 and x 2 2. Without these last two conditions, the number of solutions is ( 5+5- 1 5 ) = 126. Of these, the number with x 1 4 is ( 1+5- 1 1 ) = 5 and 1 the number with x 2 3 is ( 2+5- 1 2=15 ) . Since there are no solutions with both x 1 4 and x 2 3, then the number of solutions with...
View Full Document

This note was uploaded on 10/10/2009 for the course MATH 55 taught by Professor Strain during the Spring '08 term at University of California, Berkeley.

Page1 / 5

hw7sol - Math 55: Discrete Mathematics UC Berkeley, Spring...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online