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Question
suppose we want to choose 4 objects, without replacement, from 16 distinct objects. (a) how many ways can this be done, if the order of the choices is relevant? (b) how many ways can this be done, if the order of the choices is not relevant?
Step1: Identify permutation formula
When order is relevant, we use permutation formula $P(n,r)=\frac{n!}{(n - r)!}$, where $n = 16$ and $r=4$.
Step2: Calculate factorial values
$n!=16! = 16\times15\times14\times13\times12\times\cdots\times1$, $(n - r)!=(16 - 4)!=12!=12\times11\times\cdots\times1$. Then $P(16,4)=\frac{16!}{12!}=16\times15\times14\times13$.
Step3: Compute the result
$16\times15\times14\times13 = 43680$.
Step4: Identify combination formula
When order is not relevant, we use combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, with $n = 16$ and $r = 4$.
Step5: Calculate factorial values for combination
$n!=16!$, $r!=4!=4\times3\times2\times1$, $(n - r)!=12!$. Then $C(16,4)=\frac{16!}{4!×12!}=\frac{16\times15\times14\times13\times12!}{4\times3\times2\times1\times12!}$.
Step6: Simplify the combination expression
$\frac{16\times15\times14\times13}{4\times3\times2\times1}=1820$.
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(a) 43680
(b) 1820