Suppose we want to choose 3 letters, without replacement, from the 4 letters a, b, c, and d. (a) how many ways can this be done, if the order of the choices matters? (b) how many ways can this be done, if the order of the choices does not matter?

A) 24 ways

B) 4 ways

Step-by-step explanation:

a) permutation occurrs when order of choices matters.

N = 4P3 = 4!/(4-3)! = 4!/1!

N = 24 ways

b) combination occurs when order of choices doesn't matter.

N = 4C3 = 4!/3!(4-3)! = 4!/3!(1!)

N = 4 ways

i dont know sorry

step-by-step explanation:

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.