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I need a function that can solve the following: for a binomial function nCr=k, given r and k find n. in mathematics nCr=n!/r!(n-r)! I tried following but it doesn't solve it. for example 8C6=28, for my function the inputs are 6 and 28 and i want to find 8. This may not have exact integer number so I want to find an x>=n.

"""I am approaching it this way, i.e. find the solution of a polynomial function iteratively, hope there is a better way"""
def find_n(r,k):
    #solve_for_n_in(n*(n-1)...(n-r)=math.factorial(r)*k
    #in the above example solve_for_n(n*(n-1)(n-2)(n-3)(n-4)(n-5)=720*28)

    sum=math.factorial(r)*k
    n=r+1
    p=1

    while p<sum:
        p=1
        for i in range(0,r+2):
            p*=(n-i)
        n+=1
    return n-1

Thanks.

white_chick
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    It looks like you want us to write some code for you. While many users are willing to produce code for a coder in distress, they usually only help when the poster has already tried to solve the problem on their own. A good way to demonstrate this effort is to include the code you've written so far, example input (if there is any), the expected output, and the output you actually get (console output, tracebacks, etc.). The more detail you provide, the more answers you are likely to receive. Check the [FAQ] and [ask]. – Morgan Thrapp Mar 02 '16 at 16:46
  • I still don't know what your question is? Does this do what you want it to? If not, give specific details about what doesn't work with example input and output. – Morgan Thrapp Mar 02 '16 at 16:59
  • for example **8C6**=28, for my function the inputs are 6 and 28 and i want to find 8. – white_chick Mar 02 '16 at 17:06
  • [this may help](https://en.wikipedia.org/wiki/Binomial_coefficient#Recursive_formula) – Jared Goguen Mar 02 '16 at 23:00
  • Thanks all, I have solved it. comments and suggestion are welcome. looking for more efficient solution though. – white_chick Mar 03 '16 at 11:28

2 Answers2

0

I solved it the following way, i.e. find the solution of a polynomial function iteratively, hope there is a better way.

def find_n(r,k):
    #solve_for_n_in(n*(n-1)...(n-r)=math.factorial(r)*k
    #in the above example solve_for_n(n*(n-1)(n-2)(n-3)(n-4)(n-5)=720*28)

    target=math.factorial(r)*k
    n=r+1
    p=1

    while p<target:
        p=1
        for i in range(0,r+2):
            p*=(n-i)
        n+=1
    return n-1
white_chick
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0

Here's a solution which uses fminsearch. You'll want to minimize the absolute difference between nchoosek(n, r) and k. However, you'll likely run into undefined values for nchoosek, so it's better to define it from scratch. Don't use factorial either though, as it's undefined for negative integers. Instead, use gamma (read about this on Wikipedia if you don't know).

r = 6;
k = 28;
toMinimize = @(n) abs(gamma(n+1) / (gamma(r+1) * gamma(n-r+1)) - k);

Be smart about the initial conditions:

for n = 1:10
    [res(n, 1), fval(n, 1)] = fminsearch(toMinimize, n);
end
[res fval]

Now you'll see you should only trust initial conditions n0 >= 5, for which the answer is n = 8.

ans =
             1.42626953125          27.9929874410369
             1.42626953125          27.9929874410369
           3.5737060546875          27.9929874410073
             3.57373046875          27.9929874410369
                         8                         0
          8.00000152587891       5.2032510172495e-05
          8.00000152587891      5.20325100552554e-05
                         8                         0
          7.99999694824218      0.000104064784270719
                         8                         0
Derek
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