Solutions from a student (dragged) 22

Solutions from a student (dragged) 22 - committees and...

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

View Full Document Right Arrow Icon
Finally, if no person has more than one job, then this combination gives n ( n - 1)( n - 2)2 n - 3 total committees. Adding all of these possible combinations up we ±nd that n ( n - 1)( n - 2)2 n - 3 +3 n ( n - 1)2 n - 2 + n 2 n - 1 = n 2 ( n +3)2 n - 3 . Problem 13 (an alternating series of binomial coeFcients) From the binomial theorem we have ( x + y ) n = n ± k =0 ² n k ³ x k y n - k . If we select x = - 1and y =1then x + y =0andthesumabovebecomes 0= n ± k =0 ² n k ³ ( - 1) k , as we were asked to prove. Problem 14 (committees and subcommittees) Part (a): Pick the committee of size j in ² n j ³ ways. The subcommittee of size i from these j can be selected in ² j i ³ ways, giving a total of ² j i ³² n j
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: committees and subcommittee. Now assume that we pick the subcommittee rst. This can be done in n i ways. We then pick the committee in n-i j-i ways resulting in a total n i n-i j-i . Part (b): I think that the lower index on this sum should start at i (the smallest subcom-mittee size). If so then we have n j = i n j j i = n j = i n i n-i j-i = n i n j = i n-i j-i = n i n-i j =0 n-i j = n i 2 n-i ....
View Full Document

Ask a homework question - tutors are online