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,
[tex](\left {n} \atop {k}} \right. )=(\left {10} \atop {5}} \right. )=252[/tex]
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,
[tex](\left {n+k-1} \atop {k}} \right. )=(\left {14} \atop {5}} \right. )=2002[/tex]
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,
[tex]P\left {n} \atop {k}} \right=P\left {10} \atop {5}} \right. =30240[/tex]
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
[tex]n^k=10^5=100,000[/tex]
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.
