Prove the following two identities both algebraically and by interpreting their meaning
combinatorially.
1. (n chooses k)= (n chooses n-k)
2.(n chooses k) = ((n-1) chooses (k-1)) + ((n-1) choose k))
Solution
1. nCk = n! / (k!.(n-k)!)
nC(n-k) = n! / [(n-k)!.k!]
Hence algebraically equal.
selection of k out of n and rejection of (n-k) objects out of n is the same thing.
Hence these are equal.
2. (n-1) C (k-1) + n-1 C k = (n-1)!/[(k-1)!.(n-k)!] + (n-1)!/[k!(n-k-1)!]
= (n-1)!/[(k-1)!(n-k-1)] * (1/n-k + 1/k)
= (n-1)!/[(k-1)!(n-k-1)] * n/[(n-k).k]
= n! / k!.(n-k)!
=nCk.
Prove the following two identities both algebraically and by interpr.pdf
1. Prove the following two identities both algebraically and by interpreting their meaning
combinatorially.
1. (n chooses k)= (n chooses n-k)
2.(n chooses k) = ((n-1) chooses (k-1)) + ((n-1) choose k))
Solution
1. nCk = n! / (k!.(n-k)!)
nC(n-k) = n! / [(n-k)!.k!]
Hence algebraically equal.
selection of k out of n and rejection of (n-k) objects out of n is the same thing.
Hence these are equal.
2. (n-1) C (k-1) + n-1 C k = (n-1)!/[(k-1)!.(n-k)!] + (n-1)!/[k!(n-k-1)!]
= (n-1)!/[(k-1)!(n-k-1)] * (1/n-k + 1/k)
= (n-1)!/[(k-1)!(n-k-1)] * n/[(n-k).k]
= n! / k!.(n-k)!
=nCk
![Prove the following two identities both algebraically and by interpreting their meaning
combinatorially.
1. (n chooses k)= (n chooses n-k)
2.(n chooses k) = ((n-1) chooses (k-1)) + ((n-1) choose k))
Solution
1. nCk = n! / (k!.(n-k)!)
nC(n-k) = n! / [(n-k)!.k!]
Hence algebraically equal.
selection of k out of n and rejection of (n-k) objects out of n is the same thing.
Hence these are equal.
2. (n-1) C (k-1) + n-1 C k = (n-1)!/[(k-1)!.(n-k)!] + (n-1)!/[k!(n-k-1)!]
= (n-1)!/[(k-1)!(n-k-1)] * (1/n-k + 1/k)
= (n-1)!/[(k-1)!(n-k-1)] * n/[(n-k).k]
= n! / k!.(n-k)!
=nCk](https://image.slidesharecdn.com/provethefollowingtwoidentitiesbothalgebraicallyandbyinterpr-230326113948-58ebb592/85/Prove-the-following-two-identities-both-algebraically-and-by-interpr-pdf-1-320.jpg)
