Question

An admiral, captain, and commander, all different, are to be chosen from a group of 10 Starfleet officers. How many different choices of officers are possible if there are no restrictions?

Answer 1

Answer:

$10C1 *9C1*8C1 = 10*9*8=720$

Step-by-step explanation:

For this case in order to select the one admiral, captain and commander, all different. We are assuming that the order in the selection no matter, so we can begin selecting an admiral then a captain and then a commander.

So we have 10C1 ways to select one admiral since we want just one

Now we have remaining 9 people and we have 9C1 ways to select a captain since we want a captain different from the admiral selected first

Now we have remaining 8 people and we have 8C1 ways to select a commander since we want a commander different from the captain selected secondly.

The term nCx (combinatory) is defined as:

$nCx = \frac{n!}{x! (n-x)!}$

And by properties $nC1= n$

So then the number of possible way are:

$10C1 *9C1*8C1 = 10*9*8=720$

If we select first the captain then the commander and finally the admiral we have tha same way of select $10*9*8=720$

For all the possible selection orders always we will see that we have 720 to select.

Answer 2

Answer:

N = 10P3

N = (10×9×8×7!)/7!

N = 10×9×8 = 720 choices

Therefore, the admiral, captain, and commander can be selected in 720 ways

Step-by-step explanation:

In the case above, we need to select 3 distinct leaders from a group of 10 Officers.

This can simply be done by using permutation because im this case order is important (the three posts are distinct)

nPr = n!/(n-r)! .......1

For the case above, the number of ways we could select 3 distinct leaders from 10 Officers is;

N = 10P3

Using equation 1

N = 10!/(10-3)!

N = 10!/7!

N = (10×9×8×7!)/7!

N = 10×9×8 = 720 ways

Therefore, the admiral, captain, and commander can be selected in 720 ways