6.1 Q-3

Exercise 6.1 — Question 3: Factorial Simplification in Permutations

This question practices simplifying factorial expressions, which is essential for evaluating permutation formulas of the form:

${}^n{P}_r=\frac{n!}{(n-r)!}$

Key Property

The recursive definition of factorial: $n!=n\times (n-1)!=n(n-1)(n-2)!=n(n-1)(n-2)(n-3)!\cdots$

This allows cancellation of common factorial terms in numerator and denominator.

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Worked Examples

Part (i): Simplify $\frac{n!}{(n-2)!}$

$\frac{n!}{(n-2)!}=\frac{n(n-1)(n-2)!}{(n-2)!}=n(n-1)$

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Part (ii): Simplify $\frac{(n+1)!}{(n-1)!}$

$\frac{(n+1)!}{(n-1)!}=\frac{(n+1)\cdot n\cdot (n-1)!}{(n-1)!}=(n+1)\cdot n=n^2+n$

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Part (iii): Express $n(n-1)(n-2)$ as a factorial ratio

Multiply and divide by $(n-3)!$:

$n(n-1)(n-2)=\frac{n(n-1)(n-2)(n-3)!}{(n-3)!}=\frac{n!}{(n-3)!}$

This equals ${}^n{P}_3$.

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Part (iv): Simplify $\frac{(2n)!}{(2n-2)!}$

$\frac{(2n)!}{(2n-2)!}=\frac{(2n)(2n-1)(2n-2)!}{(2n-2)!}=(2n)(2n-1)=4n^2-2n$

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Connection to Permutations

Every permutation ${}^n{P}_r$ simplifies using the same cancellation:

${}^n{P}_r=\frac{n!}{(n-r)!}=n(n-1)(n-2)\cdots (n-r+1)(r\text{ factors})$

Mastering factorial simplification is therefore the foundation for all permutation calculations.