proof thise for me. nCr=nCr-r
2) nCr+nCr+1=n+1Crr+1
To prove the equality "nCr = nCr-r," and "nCr + nCr+1 = n+1Cr+1," we will use the factorial definition of combination (nCr) and some algebraic manipulations. The factorial notation "n!" represents the product of all positive integers less than or equal to n.
1) Proof: nCr = nCr-r
The combination formula states that nCr = n! / (r!(n-r)!).
First, let's calculate nCr:
nCr = n! / (r!(n-r)!) -- (1)
Now, let's calculate nCr-r:
nCr-r = (n - (r - r))C(r - r)
= (n - 0)C(0)
= nC0
= n! / (0!(n-0)!) -- (2)
Comparing equation (1) and equation (2), we can conclude that nCr = nCr-r.
2) Proof: nCr + nCr+1 = n+1Cr+1
To prove this, we will use the same combination formula and apply some algebraic manipulations.
Let's calculate nCr:
nCr = n! / (r!(n-r)!) -- (3)
Now, let's calculate nCr+1:
nCr+1 = n! / ((r+1)!(n-r-1)!) -- (4)
Adding equations (3) and (4):
nCr + nCr+1 = n! / (r!(n-r)!) + n! / ((r+1)!(n-r-1)!)
To add the two fractions, we need to find a common denominator. Let's multiply the first fraction by (r+1)/(r+1):
= [(r+1)/(r+1)] * n! / (r!(n-r)!) + n! / ((r+1)!(n-r-1)!)
= [(r+1)n! / ((r+1)!r!(n-r)!)] + n! / ((r+1)!(n-r-1)!)
Now, let's simplify the fractions:
= [(r+1)n!+(n!)] / ((r+1)!r!(n-r)!) + n! / ((r+1)!(n-r-1)!)
= [(r+1)n!+n!(r+1)] / ((r+1)!r!(n-r)!) + n! / ((r+1)!(n-r-1)!)
= [n!(r+1+r+1)] / ((r+1)!r!(n-r)!) + n! / ((r+1)!(n-r-1)!)
= [n!(2r+2)] / ((r+1)!r!(n-r)!) + n! / ((r+1)!(n-r-1)!)
= [n!(2r+2)] / ((r+1)!(n-r)!) + (n!r) / ((r+1)!(n-r)!)
Combining both fractions:
= [n!(2r+2) + (n!r)] / ((r+1)!(n-r)!)
= [n!(2r+2+r)] / ((r+1)!(n-r)!)
= [(n!(3r+2)] / ((r+1)!(n-r)!)
Now, let's calculate n+1Cr+1:
n+1Cr+1 = (n+1)! / ((r+1)!(n-r)!) -- (5)
Comparing equation (5) with the previous result, we can conclude that nCr + nCr+1 = n+1Cr+1.
Thus, we have proved the provided equalities.