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Prove that nCr plus nCr minus 1 equals n plus 1Cr?
nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!] = n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1} = n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]} = n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]} = (n+1)!/[r!(n+1-r)!] = n+1Cr
General formulae for the combination nCr in terms of n and r?
nCr=n!/(r!(n-r)!)
How do you find the coefficients of the terms in the binomial expansion?
The coefficient of x^r in the binomial expansion of (ax + b)^n isnCr * a^r * b^(n-r)where nCr = n!/[r!*(n-r)!]
How do you do combinations in math?
nCr=n!/r!/(n-r)!
How many mm r n 18 m?
18m=18000mm i belive :)
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