Question

Evaluate: a) n!/(n-3)!b) P(6,4)/4!​

Answer 1

Step-by-step explanation:

a).

n! is

$n(n - 1)(n - 2)(n - 3)(n - 4).........(1)$

(n-3)! is

$(n - 3)(n - 4)(n - 5)..........(1)$

If we divide the two, everything behind (n-3) will just cancel out so we would be left with only

$(n)(n - 1)(n - 2)$

So the answer to a is

$n(n - 1)(n - 2)$

b

Permutations formula of P(6,4) is

6!/(6-4)!, or 6!/2!.

$\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 360$

Then we divide by 4!

$\frac{360}{4 \times 3 \times 2 \times 1} = 15$

The answer to b is 15