The mathematical representation of 20P4 is [tex]^{20}P_4 = 20 \times 19 \times 18 \times 17[/tex]. so option C is the correct.
How many ways k things out of m different things (m ≥ k) can be chosen if order of the chosen things doesn't matter?
We can use combinations for this case,
The Total number of distinguishable things is m.
Here, Out of those m things, k things are to be chosen such that their order doesn't matter.
This can be done in total of
[tex]^mC_k = \dfrac{m!}{k! \times (m-k)!} ways.[/tex]
If the order matters, then each of those choice of k distinct items would be permuted k! times,
So, total number of choices in that case would be:
[tex]^mP_k = k! \times ^mC_k = k! \times \dfrac{m!}{k! \times (m-k)!} = \dfrac{m!}{ (m-k)!}\\\\^mP_k = \dfrac{m!}{ (m-k)!}[/tex]
This is called permutation of k items chosen out of m items (all distinct).
We have to find the mathematical representation of 20P4.
[tex]^{20}P_4[/tex]
here, m = 20
k = 4
Now substitute;
[tex]^mP_k = \dfrac{m!}{ (m-k)!}\\\\^{20}P_4 = \dfrac{20!}{ (20-4)!}\\\\^{20}P_4 = \dfrac{20!}{ (16)!}\\\\\\^{20}P_4 = \dfrac{20 \times 19 \times 18 \times 17 \times 16!}{ (16)!}\\\\\\^{20}P_4 = 20 \times 19 \times 18 \times 17[/tex]
Hence, the mathematical representation of 20P4 is [tex]^{20}P_4 = 20 \times 19 \times 18 \times 17[/tex]. so option C is the correct.
Learn more about combinations and permutations here:
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