Solutions from a student (dragged) 10

Solutions from a student (dragged) 10 - by 7 4 3 = 35...

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

View Full Document Right Arrow Icon
Problem 20 (counting the number of possible parties) Part (a): There are a total of ± 8 5 ² possible groups of friends that could attend (assuming no feuds). We have ± 2 2 ² · ± 6 3 ² sets with our two feuding friends in them, giving ± 8 5 ² - ± 2 2 ² · ± 6 3 ² =36 possible groups of friends Part (b): If two Fends must attend together we have that ± 2 2 ²± 6 3 ² if the do attend the party together and ± 6 5 ² if they don’t attend at all, giving a total of ± 2 2 ²± 6 3 ² + ± 6 5 ² =26 . Problem 21 (number of paths on a grid) ±rom the hint given that we must take four steps to the right andth rees tepsup ,wecan think of any possible path as an arraignment of the letters ”U”fo rupand“R”fo
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: by 7! 4! · 3! = 35 . Problem 22 (paths on a grid through a speciFc point) One can think of the problem of going through a speciFc point (say P ) as counting the number of paths from the start A to P and then counting the number of paths from P to the end B . To go from A to P (where P occupies the (2 , 2) position in our grid) we are looking for the number of possible unique arraignments of two “U”’s and two “R”’s, which is given by 4! 2! · 2! = 6 , possible paths. The number of paths from the point P to the point B is equivalent to the number of di²erent arraignments of two “R”’s and one “U” which is given by 3! 2! · 1! = 3 ....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online