2305(1+x)n=ΣCrxr
d(1+x)ndx=r*ΣCrxr-1
xd(1+x)ndx=rΣCrxr
dxd(1+x)ndxdx=Σ r2*Cr
Differentiating the LHS and putting x as 1 we get,
Σ r2*Cr = n*2n-1+n(n-1)*2n-2
3 Answers
23052305ΣnCk*k2
= ΣnCk(k2-k)+ ΣnCkk
= Σn(n-1)k(k-1) n-2Ck-2*k(k-1) + Σnk n-1Ck-1k
= n(n-1) Σn-2Ck-2 + n Σn-1Ck-1
=n(n-1)*2n-2 +n*2n-1
Here we used the formulas:
1)nCk = nk * n-1Ck-1 = n(n-1)k(k-1) * n-2Ck-2
And
2)ΣnCk = 2n
2305The following solution is perhaps the best way of doing the problem.
This one however was not done by me .It was a given in a book.
1)The number of ways of selecting a committee out of n members is: nCk
2)Out of the k members chosen the numbers of ways of selecting a chairman and a secretary (can be the same person) is k2.
3)Hence the number of ways of selecting a committee,a chairman and a secretary is : nCk*k2
Now lets do the same counting a bit differently,
Let us first select the chairman and a secretary.
1)If chairman≠secretary ,this can be done in n(n-1) ways and the rest of the committee can be formed in 2n-2
2)If chairman=secretary ,this can be done in n ways and the rest of the committee can be formed in 2n-1
Hence Σ r2*Cr = n*2n-1+n(n-1)*2n-2
Here we used the concept that the number of ways of selecting a committe out of n members is 2n.
This is because,
we may or may not choose the first person (2 options)
we may or may not choose the second person (2 options)
....
we may or may not choose the nth person (2 options)
Hence total number of options while choosing the committee is 2n.
