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Chapter 2: Problem 18

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

Short Answer

Expert verified
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

01

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 \[ m_2=n-m_1\]
02

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: \[ \begin{pmatrix}n\\ m_1, n - m_1 \end{pmatrix}=\frac{n!}{m_1!(n-m_1)!}. \]
03

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.

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Most popular questions from this chapter

Suppose a dic is rolled twice. What are the possible values that the following random variables can take on? (i) The maximum value to appear in the two rolls. (ii) The minimum value to appear in the two rolls. (iii) The sum of the two rolls. (iv) The value of the first roll minus the value of the second roll.

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