Solutions from a student (dragged) 22

# Solutions from a student (dragged) 22 - ³ committees and...

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

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ³ committees and subcommittee. Now assume that we pick the subcommittee ±rst. This can be done in ² n i ³ ways. We then pick the committee in ² n-i j-i ³ ways resulting in a total ² n i ³² n-i j-i ³ . 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 ² n j ³² j i ³ = n ± j = i ² n i ³² n-i j-i ³ = ² n i ³ n ± j = i ² n-i j-i ³ = ² n i ³ n-i ± j =0 ² n-i j ³ = ² n i ³ 2 n-i ....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online