Questions: Suppose we want to choose 2 letters, without replacement, from the 3 letters A, B, and C. (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?

Suppose we want to choose 2 letters, without replacement, from the 3 letters A, B, and C.
(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?

Transcript text: Suppose we want to choose 2 letters, without replacement, from the 3 letters $A, B$, and $C$. (a) How many ways can this be done, if the order of the choices matters? $\square$ (b) How many ways can this be done, if the order of the choices does not matter? $\square$

Solution

Solution Steps

Step 1: Identify the problem type

This is a permutations problem where the order of selection matters.

Step 2: Apply the permutations formula

To calculate the number of ways to choose 2 items out of 3 distinct items, the formula is: \[P(n, k) = \frac{n!}{(n-k)!}\]

Step 3: Substitute the values into the formula

\[P(3, 2) = \frac{3!}{(3-2)!}\]

Step 4: Calculate the result

\[P(3, 2) = 6\]

Final Answer:

The number of permutations is 6.

Step 1: Identify the problem type

This is a combinations problem where the order of selection does not matter.

Step 2: Apply the combinations formula

To calculate the number of ways to choose 2 items out of 3 distinct items, the formula is: \[C(n, k) = \frac{n!}{k!(n-k)!}\]

Step 3: Substitute the values into the formula

\[C(3, 2) = \frac{3!}{2!(3-2)!}\]

Step 4: Calculate the result

\[C(3, 2) = 3\]

Final Answer:

The number of combinations is 3.

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