Maximum Likelihood Estimator of parameters of multinomial distribution Suppose that 50 measuring scales made by a machine are selected at random from the production of the machine and their lengths and widths are measured. It was found that 45 had both measurements within the tolerance limits, 2 had satisfactory length but unsatisfactory width, 2 had satisfactory width but unsatisfactory length, 1 had both length and width unsatisfactory. Each scale may be regarded as a drawing from a multinomial population with density
$$
\pi_{11}^{x_{11}}
\pi_{12}^{x_{12}}
\pi_{21}^{x_{21}}(1-\pi_{11}-\pi_{12}-\pi_{21})^{x_{22}}
$$
Obtain the maximum likelihood estimates of the parameters.
I have tried this by the following way:
the likelihood function is
\begin{align}
L &amp; =L(\pi_{11},\pi_{12},\pi_{21},(1-\pi_{11}-\pi_{12}-\pi_{21})) \\[8pt]
&amp; =\prod_{i=1}^{50}[\pi_{11}^{x_{11}} \pi_{12}^{x_{12}}
\pi_{21}^{x_{21}}(1-\pi_{11}-\pi_{12}-\pi_{21})^{x_{22}}] \\[8pt]
&amp; =[\pi_{11}^{x_{11}} \pi_{12}^{x_{12}}
\pi_{21}^{x_{21}}(1-\pi_{11}-\pi_{12}-\pi_{21})^{x_{22}}]^{50} \\[8pt]
&amp; =[\pi_{11}^{45}\pi_{12}^{2}
\pi_{21}^{2}(1-\pi_{11}-\pi_{12}-\pi_{21})^{1}]^{50} \\[8pt]
&amp; =\pi_{11}^{2250}\pi_{12}^{100}
\pi_{21}^{100}(1-\pi_{11}-\pi_{12}-\pi_{21})^{50}
\end{align}
Taking logarithm of the likelihood function yields,
\begin{align}
L^*&amp; =\log L=\log \left[\pi_{11}^{2250} \pi_{12}^{100}
\pi_{21}^{100}(1-\pi_{11}-\pi_{12}-\pi_{21})^{50}\right] \\[8pt]
&amp; =2250\log [\pi_{11}]+100\log [\pi_{12}]+100\log [\pi_{21}]+50\log (1-\pi_{11}-\pi_{12}-\pi_{21})
\end{align}
Now taking the first derivative of $L^*$ with respect to $\pi_{11}$
$\frac{\partial L^*}{\partial \pi_{11}}$
$=\frac{2250}{\pi_{11}}-\frac{50}{(1-\pi_{11}-\pi_{12}-\pi_{21})}$
setting $\frac{\partial L^*}{\partial \pi_{11}}$ equal to $0$,
$$\frac{\partial L^*}{\partial \hat\pi_{11}}=0$$
$$\Rightarrow\frac{2250}{\hat\pi_{11}}-\frac{50}{(1-\hat\pi_{11}-\hat\pi_{12}-\hat\pi_{21})}=0$$
$$\Rightarrow \hat\pi_{11}=\frac{45(1-\hat\pi_{12}-\hat\pi_{21})}{44}$$
$\bullet$Are the procedure and estimate of $\pi_{11}$ correct?
$\bullet$I have another question that if it is multinomial then where the term $\binom{n}{x_{11}x_{12}x_{21}x_{22}}=\binom{50}{45,2,2,1}$?
Consider a positive integer $n$ and a set of positive real numbers $\mathbf p=(p_x)$ such that $\sum\limits_xp_x=1$. The multinomial distribution with parameters $n$ and $\mathbf p$ is the distribution $f_\mathbf p$ on the set of nonnegative integers $\mathbf n=(n_x)$ such that $\sum\limits_xn_x=n$ defined by
$$
f_\mathbf p(\mathbf n)=n!\cdot\prod_x\frac{p_x^{n_x}}{n_x!}.
$$
For some fixed observation $\mathbf n$, the likelihood is
$L(\mathbf p)=f_\mathbf p(\mathbf n)$ with the constraint $C(\mathbf p)=1$, where $C(\mathbf p)=\sum\limits_xp_x$. To maximize $L$, one asks that the gradient of $L$ and the gradient of $C$ are colinear, that is, that there exists $\lambda$ such that, for every $x$, 
$$
\frac{\partial}{\partial p_x}L(\mathbf p)=\lambda\frac{\partial}{\partial p_x}C(\mathbf p).
$$
In the present case, this reads
$$
\frac{n_x}{p_x}L(\mathbf p)=\lambda,
$$
that is, $p_x$ should be proportional to $n_x$. Since $\sum\limits_xp_x=1$, one gets finally $\hat p_x=\dfrac{n_x}n$ for every $x$.
We can then say: (i.e., Lemma)
If $n_i&gt;0$, for every $i$, and subject to $0&lt;p_i&lt;1$. Then $$f(p_1,p_2,..., p_q)= \sum^q_{i=1} n_i \log_e{p_i} $$  ...  the only expression with  $p_i 's $ we are interested in the reduced  log likelihood (p).
So, the maximum of $f(p_1,p_2,..., p_q)$ is achieved at the points $(\hat{p_i}, \hat{p_i}, ... , \hat{p_q})$ where $\hat{p_i}= \frac{n_i}{n} $ for every i.