Question

1. how many combinations with repetition are allowed if n = 7 and r = 2?

36
28
14
49

Explanation:

Step1: Recall the formula for combinations with repetition

The formula for combinations with repetition is $C(n + r - 1, r)=\frac{(n + r - 1)!}{r!(n - 1)!}$, where $n = 7$ and $r = 2$.

Step2: Substitute the values into the formula

Substitute $n = 7$ and $r = 2$ into the formula:
First, calculate $n + r - 1=7 + 2 - 1 = 8$.
Then, $C(8, 2)=\frac{8!}{2!(8 - 2)!}=\frac{8!}{2!6!}$.
Since $n!=n\times(n - 1)\times\cdots\times1$, we have $8! = 8\times7\times6!$ and $2! = 2\times1$.
So, $\frac{8\times7\times6!}{2\times1\times6!}=\frac{8\times7}{2\times1}$.
Calculate $8\times7 = 56$ and $2\times1 = 2$. Then $\frac{56}{2}=28$.

Answer:

28