Question

You are given the following letters. A. B. C. D. E. F. G. H. If you are asked to randomly choose 3 of the letters and lay it out in a sequence (order matters), what is the total number of outcomes?

a) 84
b) 168
c) 336
d) 504

Answer 1

To find the total number of outcomes when randomly choosing 3 letters and laying them out in a sequence, we can use permutations. In this case, there are 8 letters to choose from and we are choosing 3, resulting in a total of 336 outcomes.

Step-by-step explanation:

To find the total number of outcomes when randomly choosing 3 letters and laying them out in a sequence, we need to use permutations. The formula for permutations is nPr = n! / (n-r)!, where n is the total number of items and r is the number of items being chosen.

In this case, there are 8 letters to choose from and we are choosing 3. So the number of outcomes is 8P3 = 8! / (8-3)! = 8! / 5! = 8 x 7 x 6 = 336.

Therefore, the total number of outcomes is 336, which corresponds to option c.