Given that 𝑛𝐶4𝑛𝑃2= 12, find n

The given equation is 𝑛𝐶4𝑛𝑃2= 12.

We know that 𝑛𝐶4 = n! / [(n-4)! * 4!] and 𝑛𝑃2 = n! / (n-2)!.

Substituting these into the equation, we get:

(n! / [(n-4)! * 4!]) * (n! / (n-2)!) = 12

Simplifying, we get:

(n! * n!) / [(n-4)! * 4! * (n-2)!] = 12

We can cancel out one n! from the numerator and denominator, and we get:

n! / [(n-4)! * 4! * (n-2)] = 12

We know that (n-2) = (n-4+2), so we can rewrite the equation as:

n! / [(n-4)! * 4! * (n-4+2)] = 12

Solving for n, we get:

n = 6.