Suppose we want to choose 2 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?

Letters can be chosen in 12 different ways, if order matters, or 6 different ways, if order doesn't matter.

Step-by-step explanation:

Since we want to choose 2 letters, without replacement, from the 4 letters A, B, C, and D, to determine in how many ways can this be done, if the order of the choices matters, and in how many ways can this be done, if the order of the choices does not matter, the following calculations must be performed:

If order matters =

(4 x 3 x 2 x 1) / 2 = X

24/2 = X

12 = X

If the order doesn't matter =

12/2 = X

6 = X

Therefore, letters can be chosen in 12 different ways, if order matters, or 6 different ways, if order doesn't matter.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

21/58.

hope to be marked brainliest