Question

Show that when \(r=2\) the multinomial reduces to the binomial.

Short answer

When \(r=2\), the multinomial theorem becomes \(\begin{pmatrix}n\\ m_1, m_2 \end{pmatrix}=\frac{n!}{m_1!m_2!}\) with \(m_2=n-m_1\). Replacing \(m_2\) with \(n-m_1\) gives us \(\begin{pmatrix}n\\ m_1, n - m_1 \end{pmatrix}=\frac{n!}{m_1!(n-m_1)!}\), which is equivalent to the binomial theorem \(\binom{n}{m_1}=\frac{n!}{m_1!(n-m_1)!}\). Hence, when \(r=2\), the multinomial reduces to the binomial.

Step-by-step solution

Input \(r=2\) into the Multinomial Theorem

When we input \(r=2\) into the multinomial theorem, we have two types of events (\(m_1\) and \(m_2\)). The multinomial becomes: \[ \begin{pmatrix}n\\ m_1, m_2 \end{pmatrix}=\frac{n!}{m_1!m_2!}. \] Since we have only 2 types of events and the total number of events is \(n\), we can express the relationship between \(m_1\) and \(m_2\) as \[ m_2=n-m_1\]

Replace \(m_2\) with \(n - m_1\)

Now that we have expressed the relationship between \(m_1\) and \(m_2\), we will replace \(m_2\) with \(n - m_1\) in the multinomial: \[ \begin{pmatrix}n\\ m_1, n - m_1 \end{pmatrix}=\frac{n!}{m_1!(n-m_1)!}. \]

Show Equivalence to the Binomial Theorem

Now let's compare our reduced multinomial to the binomial theorem. We can rewrite the binomial theorem with \(k = m_1\): \[\binom{n}{m_1}=\frac{n!}{m_1!(n-m_1)!}.\] Thus, we have: \[ \begin{pmatrix}n\\ m_1, n - m_1 \end{pmatrix}=\frac{n!}{m_1!(n - m_1)!} = \binom{n}{m_1}. \] So we have shown that when \(r=2\), the multinomial reduces to the binomial.