Question

1. solve each equation, where ( n ) is an integer.

( \boldsymbol{\text{a)}} ) ( \frac{n!}{84} = {}_{n - 2}p_{n - 4} )
( \boldsymbol{\text{b)}} ) ( {}_np_4 = 8({}_{n - 1}p_3) )

Explanation:

Part (a)

Step 1: Recall Permutation Formula

The permutation formula is \( _nP_r=\frac{n!}{(n - r)!} \). Given \( \frac{n!}{84}=_{n - 2}P_{n - 4} \), first find the valid range for \( n \). For permutation \( _aP_b \), we need \( a\geq b\geq0 \) and \( a\in\mathbb{Z}^+ \), \( b\in\mathbb{Z}^+ \cup\{0\} \). So for \( _{n - 2}P_{n - 4} \), we have \( n-2\geq n - 4\geq0 \). From \( n - 4\geq0 \), we get \( n\geq4 \). Also, \( n-2\geq n - 4 \) is always true for \( n\in\mathbb{R} \). Now, use the permutation formula for \( _{n - 2}P_{n - 4} \):
\( _{n - 2}P_{n - 4}=\frac{(n - 2)!}{(n - 2-(n - 4))!}=\frac{(n - 2)!}{2!}=\frac{(n - 2)!}{2} \)
So the equation becomes \( \frac{n!}{84}=\frac{(n - 2)!}{2} \)

Step 2: Simplify \( n! \)

We know that \( n!=n\times(n - 1)\times(n - 2)! \). Substitute this into the equation:
\( \frac{n\times(n - 1)\times(n - 2)!}{84}=\frac{(n - 2)!}{2} \)
Since \( (n - 2)!
eq0 \) (because \( n\geq4 \)), we can divide both sides by \( (n - 2)! \):
\( \frac{n(n - 1)}{84}=\frac{1}{2} \)

Step 3: Solve for \( n \)

Cross - multiply: \( 2n(n - 1)=84 \)
Simplify: \( n(n - 1)=42 \)
Expand: \( n^2 - n - 42 = 0 \)
Factor the quadratic equation: \( n^2 - n - 42=(n - 7)(n + 6)=0 \)
So \( n = 7 \) or \( n=-6 \). But since \( n\geq4 \) (from permutation definition), we discard \( n=-6 \).

Part (b)

Step 1: Recall Permutation Formula

Using the permutation formula \( _nP_r=\frac{n!}{(n - r)!} \), for \( _nP_4 \) and \( _{n - 1}P_3 \), we have:
\( _nP_4=\frac{n!}{(n - 4)!} \) and \( _{n - 1}P_3=\frac{(n - 1)!}{(n - 1 - 3)!}=\frac{(n - 1)!}{(n - 4)!} \)
The equation is \( _nP_4 = 8\times_{n - 1}P_3 \)

Step 2: Substitute Permutation Formulas

Substitute the formulas into the equation:
\( \frac{n!}{(n - 4)!}=8\times\frac{(n - 1)!}{(n - 4)!} \)
Since \( (n - 4)!
eq0 \) (for \( n\geq4 \), because \( n-4\geq0 \) for permutation \( _nP_4 \) and \( _{n - 1}P_3 \)), we can divide both sides by \( (n - 4)! \):
\( n!=8\times(n - 1)! \)

Step 3: Simplify \( n! \)

We know that \( n!=n\times(n - 1)! \). Substitute this into the equation:
\( n\times(n - 1)!=8\times(n - 1)! \)
Since \( (n - 1)!
eq0 \) (for \( n\geq2 \), and from permutation \( _nP_4 \), \( n\geq4 \)), we can divide both sides by \( (n - 1)! \):
\( n = 8 \)

Answer:

s:
a) \( \boldsymbol{n = 7} \)

b) \( \boldsymbol{n = 8} \)