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?

Question

Zion Harper

Mathematics

30 August, 04:22

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?

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Answers (1)

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Arnav Baird · Answer 1

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