Questions: Suppose we want to choose 2 letters, without replacement, from the 4 letters A, B, C, and D. (If necessary, consult a list of formulas.) (a) How many ways can this be done, if the order of the choices is taken into consideration? (b) How many ways can this be done, if the order of the choices is not taken into consideration?

Solution Steps

We are tasked with determining the number of ways to choose 2 letters from the set {A, B, C, D} under two different conditions: (a) when the order of selection matters, and (b) when the order does not matter.

When the order of selection is taken into consideration, we are dealing with permutations. The formula for permutations of \( n \) items taken \( r \) at a time is given by: \[ P(n, r) = \frac{n!}{(n - r)!} \] Here, \( n = 4 \) and \( r = 2 \). Plugging in the values: \[ P(4, 2) = \frac{4!}{(4 - 2)!} = \frac{4 \cdot 3 \cdot 2!}{2!} = 4 \cdot 3 = 12 \] So, there are 12 possible ordered pairs.

When the order of selection does not matter, we are dealing with combinations. The formula for combinations of \( n \) items taken \( r \) at a time is given by: \[ C(n, r) = \frac{n!}{r!(n - r)!} \] Again, \( n = 4 \) and \( r = 2 \). Plugging in the values: \[ C(4, 2) = \frac{4!}{2!(4 - 2)!} = \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2!} = \frac{12}{2} = 6 \] So, there are 6 possible unordered pairs.

Final Answer