(a) If the order of the choices matters, we are dealing with permutations. The formula for permutations is nPr = n! / (n-r)!, where n is the total number of options, r is the number of options chosen, and "!" denotes factorial. In this case, n = 3 (A, B, C) and r = 2 (we are choosing 2 letters). So, the calculation would be:

3P2 = 3! / (3-2)!
= 3*2*1 / 1
= 6 ways

(b) If the order of the choices does not matter, we are dealing with combinations. The formula for combinations is nCr = n! / [r!(n-r)!]. In this case, n = 3 and r = 2. So, the calculation would be:

3C2 = 3! / [2!(3-2)!]
= 3*2*1 / [2*1 * 1]
= 3 ways

So, there are 6 ways to choose 2 letters from A, B, and C if the order matters, and 3 ways if the order does not matter.

Question

(a) If the order of the choices matters, we are dealing with permutations. The formula for permutations is nPr = n! / (n-r)!, where n is the total number of options, r is the number of options chosen, and "!" denotes factorial. In this case, n = 3 (A, B, C) and r = 2 (we are choosing 2 letters). So, the calculation would be:

3P2 = 3! / (3-2)!
= 3*2*1 / 1
= 6 ways

(b) If the order of the choices does not matter, we are dealing with combinations. The formula for combinations is nCr = n! / [r!(n-r)!]. In this case, n = 3 and r = 2. So, the calculation would be:

3C2 = 3! / [2!(3-2)!]
= 3*2*1 / [2*1 * 1]
= 3 ways

So, there are 6 ways to choose 2 letters from A, B, and C if the order matters, and 3 ways if the order does not matter.

Knowee AI · Accepted Answer

(a) If the order of the choices matters, we are dealing with permutations. The formula for permutations is nPr = n! / (n-r)!, where n is the total number of options, r is the number of options chosen, and "!" denotes factorial. In this case, n = 3 (A, B, C) and r = 2 (we are choosing 2 letters). So, the calculation would be:

3P2 = 3! / (3-2)!
= 3*2*1 / 1
= 6 ways

(b) If the order of the choices does not matter, we are dealing with combinations. The formula for combinations is nCr = n! / [r!(n-r)!]. In this case, n = 3 and r = 2. So, the calculation would be:

3C2 = 3! / [2!(3-2)!]
= 3*2*1 / [2*1 * 1]
= 3 ways

So, there are 6 ways to choose 2 letters from A, B, and C if the order matters, and 3 ways if the order does not matter.

Suppose we want to choose 2 letters, without replacement, from the 3 letters A, B, and C.(If necessary, consult a list of formulas.)(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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