prove:1) nCr=nCr-r
2) nCr+nCr+1=n+1Crr+1
To prove the given identities, we will use the definition of the binomial coefficient and combinatorial reasoning.
1) Proof for nCr = nCr-r:
The binomial coefficient (nCr) represents the number of ways to choose r items from a set of n distinct items, where order does not matter. It can be expressed as:
nCr = n! / (r!(n-r)!)
Using the definition of factorial (!), we can rewrite the equation as:
nCr = n! / ((n-r)!(r!))
Now, let's substitute (n-r) for r in the equation:
nC(n-r) = n! / ((n-(n-r))!((n-r)!))
Simplifying further:
nC(n-r) = n! / (r!(n-r)!)
Which is equal to nCr.
Therefore, we have proved that nCr = nCr-r.
2) Proof for nCr + nCr+1 = n+1Cr+1:
The binomial coefficient (nCr) represents the number of ways to choose r items from a set of n distinct items. The identity to prove states that the sum of nCr and nCr+1 is equal to (n+1)Cr+1.
Using the same definition of the binomial coefficient, we have:
nCr = n! / (r!(n-r)!)
nCr+1 = n! / ((r+1)!(n-r-1)!)
Let's find a common denominator for the two terms:
nCr x (r+1)! / (r+1)! = n!(n-r)!(r+1)! / (r!(n-r)!)(r+1)!
nCr+1 = n!(n-r)!(r+1)! / (r!(n-r)!(r+1)!)
Now, we can add the two terms:
nCr + nCr+1 = n!(n-r)!(r+1)! / (r!(n-r)!) + n!(n-r)!(r+1)! / (r!(n-r)!(r+1)!)
Combining the terms over a common denominator:
nCr + nCr+1 = (n!(n-r)!(r+1)! + n!(n-r)!(r+1)!)/ (r!(n-r)!(r+1)!)
Simplifying the numerator:
nCr + nCr+1 = (n!(n-r)!(r+1) + n!(n-r)!(r+1))/ (r!(n-r)!(r+1)!)
Canceling out the common terms:
nCr + nCr+1 = (n!(n-r)!(r+1) + n!(n-r)!(r+1))/ (r!(n-r)!(r+1)!)
= n!(n-r)!(r+1) / (r!(n-r)!(r+1)!)
Cancelling out the common terms again:
nCr + nCr+1 = (n!(n-r)!(r+1) + n!(n-r)!(r+1))/ (r!(n-r)!(r+1)!),
= ((n+1)!)/(r+1)!(n-r)!(n+1-r)!
Which is equal to (n+1)Cr+1.
Therefore, we have proved that nCr + nCr+1 = n+1Cr+1.
By following the steps above, we have proven both identities.