Answer:
a)
![N= (5C1) *(5C1) *(5C1) = 5*5*5 = 125](https://img.qammunity.org/2021/formulas/mathematics/college/j5x32h94p1yfoi1734btsu015n97frvnk2.png)
b)
![N = (5C1)*(4C1) *(3C1) = 5*4*3 = 60](https://img.qammunity.org/2021/formulas/mathematics/college/ls7kv8crkym34on1tghenl0ejvaanhj1qm.png)
Explanation:
For this case our sample space is the 5 letters given:
![S= [F,G, H, I, J]](https://img.qammunity.org/2021/formulas/mathematics/college/oht4uvbx6mdhxmjkganv4j6bheac1v3dy4.png)
And we want to find the number of three letter words can be made from the sample space with some conditions
Part a
For this case the repetition is allowed so then each time we will have 5 possibilites in order to select one letter so if we use combinatories we have:
![N= (5C1) *(5C1) *(5C1) = 5*5*5 = 125](https://img.qammunity.org/2021/formulas/mathematics/college/j5x32h94p1yfoi1734btsu015n97frvnk2.png)
So then we will have 125 possible combinations of 3 words letters with the 5 provided
We need to remember that
![nC x = (n!)/((n-x)! x!)](https://img.qammunity.org/2021/formulas/mathematics/college/8ygle9zcc4fqlkfso8aqtkvyjuz3oic2gl.png)
Part b
For this case the repetition is not allowed so then the possible number of possibilities are:
![N = (5C1)*(4C1) *(3C1) = 5*4*3 = 60](https://img.qammunity.org/2021/formulas/mathematics/college/ls7kv8crkym34on1tghenl0ejvaanhj1qm.png)
So then we will have 60 possible combinations of 3 words letters with the 5 provided