Answer:
Step-by-step explanation:
I will illustrate this solution with a unique birthday party situation between Jane (the celebrant) and her 10 friends.
In the said party , we will assume that Jane only has 5 chocolates to share among her friends
i. Assuming that the 5 chocolates are of the same type, if she doesn't want to give any friend more than one piece of chocolate, the situation here is said to be WITHOUT REPLACEMENT & UN-ORDERED
n = 10 and k = 5
Total number of possible combinations becomes,
ii. Assuming that the 5 pieces of chocolate are of the same type and she is willing to give a friend more than one piece of chocolate, the situation is said to be WITH REPLACEMENT & UN-ORDERED
n = 10 and k = 5
Total number of possible combinations becomes,
![(\left {n+k-1} \atop {k}} \right. )=(\left {14} \atop {5}} \right. )=2002](https://tex.z-dn.net/?f=%28%5Cleft%20%7Bn%2Bk-1%7D%20%5Catop%20%7Bk%7D%7D%20%5Cright.%20%29%3D%28%5Cleft%20%7B14%7D%20%5Catop%20%7B5%7D%7D%20%5Cright.%20%29%3D2002)
iii. Assuming that the 5 pieces of chocolate are of different types and she isn't willing to give any friend more than one piece of chocolate, the situation is said to be WITHOUT REPLACEMENT & ORDERED
n = 10 & k = 5
Total number of possible combinations becomes,
![P\left {n} \atop {k}} \right=P\left {10} \atop {5}} \right. =30240](https://tex.z-dn.net/?f=P%5Cleft%20%7Bn%7D%20%5Catop%20%7Bk%7D%7D%20%5Cright%3DP%5Cleft%20%7B10%7D%20%5Catop%20%7B5%7D%7D%20%5Cright.%20%3D30240)
iv. Assuming the the 5 pieces of chocolate are of different types, if she is willing to give a friend more than one piece of chocolate, the situation is said to be WITH REPLACEMENT & ORDERED
n = 10 AND k = 5
Total number of combinations becomes
![n^k=10^5=100,000](https://tex.z-dn.net/?f=n%5Ek%3D10%5E5%3D100%2C000)
Sampling with replacement means that one friend can be sampled more than once i.e: A friend receives a piece of chocolate more than once
The order dictates how the sampling is applied.