Question

1. eight groups in a chemistry class submit their own drawing of a molecule for a science fair project.

a. four of the groups will be randomly selected to participate in the science fair. in how many ways can the teacher select the groups?
type of problem:

b. three groups attend the science fair, but a member of the fourth group gets sick so the group cannot attend. how many ways can the remaining groups be chosen to replace the absent group?
type of problem:

c. the four groups will have their drawings taped up on a wall in a hallway. in how many ways can these drawings be reordered from left to right?
type of problem:

1. determine and justify whether the statement is always, sometimes, or never true: $_nc_r = _np_r$. write at least two sentences and include at least two examples to explain your answer.

a. 720 ways
70 ways
1b. same
4b. 4 ways

1. 3125 patterns

4c. 24 ways

1. 150 ratings
2. sometimes

Explanation:

Step1: Identify combination formula

The formula for combinations is $_{n}C_{r}=\frac{n!}{r!(n - r)!}$, where $n$ is the total number of items and $r$ is the number of items to be chosen. For part a, $n = 8$ and $r=4$.
$_{8}C_{4}=\frac{8!}{4!(8 - 4)!}$

Step2: Calculate factorial values

$8! = 8\times7\times6\times5\times4\times3\times2\times1=40320$, $4! = 4\times3\times2\times1 = 24$, and $(8 - 4)!=4!=24$. Then $_{8}C_{4}=\frac{8\times7\times6\times5\times4!}{4!\times4!}=\frac{8\times7\times6\times5}{4\times3\times2\times1}=70$.

Step3: Analyze part b

One group is out, so there are $8- 3=4$ remaining groups to choose from to replace the absent - group. So there are 4 ways.

Step4: Identify permutation formula for part c

The formula for permutations of $n$ distinct objects is $n!$. Here $n = 4$, so the number of ways to re - order the 4 drawings is $4!$.
$4!=4\times3\times2\times1 = 24$.

Step5: Analyze statement in part 5

The combination formula is $_{n}C_{r}=\frac{n!}{r!(n - r)!}$ and the permutation formula is $_{n}P_{r}=\frac{n!}{(n - r)!}$. When $r = 0$ or $r=1$, $_{n}C_{r}=_{n}P_{r}$. For example, when $n = 5$ and $r = 1$, $_{5}C_{1}=\frac{5!}{1!(5 - 1)!}=\frac{5!}{1!4!}=5$ and $_{5}P_{1}=\frac{5!}{(5 - 1)!}=\frac{5!}{4!}=5$. When $n=5$ and $r = 2$, $_{5}C_{2}=\frac{5!}{2!(5 - 2)!}=\frac{5\times4\times3!}{2\times1\times3!}=10$ and $_{5}P_{2}=\frac{5!}{(5 - 2)!}=\frac{5\times4\times3!}{3!}=20$. So the statement $_{n}C_{r}=_{n}P_{r}$ is sometimes true.

Answer:

a. 70 ways
b. 4 ways
c. 24 ways
d. Sometimes