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Prove that for all integers n ≥ 3, n+1P3 −n P3 = 3 · n P2

Sagot :

MrRoyal

The permutation equation [tex]^{n+1}P_3 - ^nP_3= 3 * ^nP_2[/tex] is true for all integers n ≥ 3

How to prove the equation?

We have:

[tex]^{n+1}P_3 - ^nP_3= 3 * ^nP_2[/tex]

Apply the following permutation formula

[tex]^{n}P_r = \frac{n!}{(n- r)!}[/tex]

So, we have:

[tex]\frac{(n + 1)!}{(n + 1 - 3)!} - \frac{n!}{(n - 3)!} = 3 * \frac{n!}{(n - 2)!}[/tex]

Evaluate the difference

[tex]\frac{(n + 1)!}{(n - 2)!} - \frac{n!}{(n - 3)!} = 3 * \frac{n!}{(n - 2)!}[/tex]

Expand

[tex]\frac{(n + 1)!}{(n - 2)(n - 3)!} - \frac{n!}{(n - 3)!} = 3 * \frac{n!}{(n - 2)(n - 3)!}[/tex]

Multiply through by (n - 3)!

[tex]\frac{(n + 1)!}{(n - 2)} - n! = 3 * \frac{n!}{(n - 2)}[/tex]

Expand

[tex]\frac{(n + 1) * n!}{(n - 2)} - n! = 3 * \frac{n!}{(n - 2)}[/tex]

Divide through by n!

[tex]\frac{(n + 1)}{(n - 2)} - 1 = \frac{3}{(n - 2)}[/tex]

Take the LCM

[tex]\frac{n + 1 - n + 2}{(n - 2)} = \frac{3}{(n - 2)}[/tex]

Evaluate the like terms

[tex]\frac{3}{(n - 2)} = \frac{3}{(n - 2)}[/tex]

Both sides of the equations are the same.

Hence, the permutation equation [tex]^{n+1}P_3 - ^nP_3= 3 * ^nP_2[/tex] is true for all integers n ≥ 3

Read more about permutation at:

https://brainly.com/question/11732255

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