quiz7 - 2 n n = n n n + n 1 n n-1 + n 2 n n-2 + + n n n...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Name: Math 55 Quiz 7 SOLUTIONS July 22, 2009 GSI: Rob Bayer You have until 4:00 to complete this quiz. You must show your work. 1. (3 pts) In poker, a “straight” consists of five cards with consecutive ranks and of any suits (8 , 9 , 10 ,J ,Q for example). Assume Aces are considered higher than Kings and straights may not “wrap around” from Aces back to 2’s. How many ways are there to make a straight from a standard 52-card deck? Be sure to at least briefly justify your answer. 9 |{z} starting rank · 4 5 |{z} suits 2. (3 pts) Find the number of solutions in non-negative integers to x 1 + x 2 + x 3 + x 4 + x 5 = 55 with x 2 5 This is the same as the number of solutions to x 1 + x 2 + x 3 + x 4 + x 5 = 50, which we know is ± 50 + 5 - 1 5 - 1 ² = ± 54 4 ² 3. (4 pts) Prove that
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 n n = n n n + n 1 n n-1 + n 2 n n-2 + + n n n Note that this is a special case of Vandermondes Identity, and you may only use that theorem if you prove it here. Consider two sets, S and T that are disjoint and each have n elements in them. Then the LHS counts the number of ways to choose n elements from S T . We can also count this same quantity as follows: For each k between 0 and n , there are ( n k )( n n-k ) ways to choose k elements from S and the remaining n-k elements from T . Thus, the total number of ways is also X n k n n-k which is exactly the RHS....
View Full Document

This note was uploaded on 08/31/2009 for the course MATH 55 taught by Professor Strain during the Summer '08 term at University of California, Berkeley.

Ask a homework question - tutors are online