Hey all, I'm a bit stuck at verifying these following formulas are equivalent:

Prove:

(n Choose k) = (n choose n-k)

so I got:

= n!/k!(n-k)! = n!/(n-k)!(n-n-k)!

= n!/k!(n-k)! = n!/(n-k)!(-k)!

Is the k on the right hand side supposed to be negative??

(n Choose k)= n!/[k!(n-k)!]

(n choose n-k)=n!/[(n-k)!(n-(n-k))!]=n!/[(n-k)!(n-n+k)!]=n!/[(n-k)!k!]

=n!/[k!(n-k)!]

Also, I have another:

Prove:

(n Choose K) = (n-1 Choose k-1) + (n-1 Choose k)

Any pointers would be helpful!

James

(n-1 Choose k-1) + (n-1 Choose k)

=(n-1)!/[(k-1)!((n-1)-(k-1))!]+(n-1)!/[k!((n-1)-k)!]

=(n-1)!/[(k-1)!(n-k)!]+(n-1)!/[k!(n-k-1)!]

multiply through by n/n

=n/n*{(n-1)!/[(k-1)!(n-k)!]+(n-1)!/[k!(n-k-1)!]}

=n!*[1/(n*(k-1)!(n-k)!)+1/(n*k!(n-k-1)!)]

=n!*[(n*k!(n-k-1)!)+(n*(k-1)!(n-k)!)]/[(n*(k-1)!(n-k)!)*(n*k!(n-k-1)!)]

=n!*[k!(n-k-1)!)+(k-1)!(n-k)!]/[n*(k-1)!(n-k)!*k!(n-k-1)!]

=n!*[(k-1)!(n-k-1)!(k+(n-k))]/[n*(k-1)!(n-k)!*k!(n-k-1)!]

=n!*[(k-1)!(n-k-1)!n]/[n*(k-1)!(n-k)!*k!(n-k-1)!]

=n!*[1]/[(n-k)!*k!]

=n!/[(n-k)!*k!]=nCk