Question

If n! = 120× (n-3), then find the value of n.
pls give a correct answer.
or else I will Lodge a complaint.​

Answer 1

Answer:

n = 6

Step-by-step explanation:

I believe your Question is,

"If n! = 120 × (n - 3)!, then find the value of n."

Now, we have

n! = 120 × (n - 3)!

n!/(n - 3)! = 120

We know that,

n! = n(n - 1)(n - 2)(n - 3)!

So,

n(n - 1)(n - 2)(n - 3)!/(n - 3)! = 120

Cancelling out (n - 3)! we get,

n(n - 1)(n - 2) = 120

Now, we should find 3 consecutive numbers that will give 120 as their product.

120 = 2 × 2 × 2 × 3 × 5

Using Associative Property,

120 = (2 × 2) × (2 × 3) × 5

120 = 4 × 5 × 6

Thus,

n(n - 1)(n - 2) = 6 × 5 × 4

Hence,

n = 6

OR

This is a bit long.......

n(n - 1)(n - 2) = 120

n(n² - 2n - n + 2) = 120

n(n² - 3n + 2) = 120

n³ - 3n² + 2n = 120

n³ - 3n² + 2n - 120 = 0

Solving this is a bit difficult

This Quadratic equation has 1 real root and 2 complex roots.

Solving this we will get the real root as n = 6.

Hence,

n = 6

Hope it helped and believing you understood it........All the best

Answer 2

Answer:

n-6

Step-by-step explanation:

n= 6 please mark me as a brainliest