Discrepancy between example using the binomial distribution and binomial probability mass function definition

Question

When studying likelihoods, I was presented with the following example:

A drug is given to $n$ patients. We observe $y_1, \dots, y_n$ where $y_i = 1$ if the $i$th patient is cured and $0$ otherwise. Our model may be that $y_i$ is a realisation of $Y_i$, and $Y_i$ has a binomial distribution with parameter $p$; and that the $Y_i$ are independent of each other.

The solution is provided as follows:

Assume $n = 20$ and $\sum_{i = 1}^n y_i = 12$. We give the plot of $x^{\sum_{i = 1}^n y_i}(1 - x)^{n - \sum_{i = 1}^n y_i}$

The R code provided is as follows:

n <- 20
sumy <- 12
curve(x^sumy * (1 - x)^(n - sumy), from = 0, to = 1, n = 301, xlab = "p", ylab = "L(p; y)")

The binomial PMF is $$Pr(k; n, p) = Pr(X = k) = {n \\ k \choose} p^k (1 - p)^{n - k}.$$

There seems to be a discrepancy between these two, where the ${n \\ k \choose}$ term is absent in the example solution binomial PMF. Why is the ${n \\ k \choose}$ term absent in the example solution binomial PMF?

Answer 1

Presumably, this code is meant to show that the maximum likelihood estimator is 12/20. To answer your question, the term ${n \choose k}$ does not affect the maximization of the function since it only modifies the likelihood by a constant of proportionality that does not depend on the unknown parameter. Thus, the point on the x-axis where the maximum occurs is unchanged by this constant. To see it analytically, differentiate the log-likelihood with respect to $p$ and you will see that $\log{n \choose k}$ drops out.