Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# FACTORIAL QUESTION?

Anyone know explain why:

n! / (n-2)! = 42

is equal to

n^2 = n + 42

Relevance

n! / (n-2)! = 42

n(n -1)(n - 2)! / (n - 2)! = 42

n(n - 1) = 42

n² - n = 42

n² = n + 42

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• n! = n(n-1)(n-2)! Therefore n!/(n-2)! = n(n-1)(n-2)!/(n-2)! = n(n-1). Then n!/(n-2)! = 42

becomes n(n-1) = 42, ie., n^2 -n -42 = 0, ie., (n+6)(n-7) = 0. Negative root n = -6 is

extraneous since factorials, by definition, exist only for non-negative integers. Then

n = 7.

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• n! / (n-2)! = 42

Alternate form assuming n is real:

n^2 = n + 42

Alternate form:

(n - 1) n = 42

Solution:

n = 7

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• n! = n x (n-1) x (n - 2) x (n - 3) x (n - 4) x...

(n - 2)! = (n - 2) x ( n-3) x ( n - 4) x ....

By cancelling down the LHS you are left with

n x ( n- 1) = 42

n^2 - n = 42

n^2 = n + 42

Hope that helps!!!!

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• n!/(n - 2)! = n(n – 1) * (n - 2)!/(n - 2)! = n^2 – n = 42

then n^2 = n + 42 or equivalently

n^2 - n – 42 = (n – 7)(n + 6) = 0

Positive option is n = 7

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• Before starting: example

n! = a * b * c * d * e * f * g ← you can see that: n = g

(n - 2)! = a * b * c * d * e

So you can deduce that:

n! = a * b * c * d * e * f * g

n! = (a * b * c * d * e) * f * g

n! = (n - 2)! * f * g → recall: g = n

n! = (n - 2)! * f * n → so, you can say that: f = n - 1

n! = (n - 2)! * (n - 1) * n

Now, the equation:

n! / (n - 2)! = 42 → recall the previous result

(n - 2)! * (n - 1) * n / (n - 2)! = 42 → you can sylpmly by (n - 2)!

n * (n - 1) = 42

n² - n = 42

n² - n + (1/2)² = 42 + (1/2)²

n² - n + (1/2)² = (168/4) + (1/4)

[n - (1/2)]² = 169/4

n - (1/2) = ± 13/2

n = (1/2) ± (13/2)

n = (1 ± 13)/2 → only the positive value

n = (1 + 13)/2

n = 7

Check it:

7! = 1 * 2 * 3 * 4 * 5 * 6 * 7 = 5040

(7 - 5)! = 5! = 120

= 5040/120

= 42

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• That depends on what I is....nI=42(n-2)I

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