Question

The first task of the Segment Two Honors Project is to select the Power Pill study participants’ groups and research board officers. Part 1: There are 40 volunteers for the research study on the Power Pill. Each subgroup of the study will contain 10 participants. Determine how many ways these participants can be selected and explain your method. Part 2: There are 15 research doctors participating in the study and the research board needs to be established with the offices of director, assistant director, quality control analyst, and correspondent. (Doctors can only hold one office on the research board.) Determine how many ways this research board can be chosen and explain your process.

Answer 1

Answer:

Part 1: There are 4.7*10^21 ways to select 40 volunteers in subgroups of 10

Part 2: The research board can be chosen in 32760 ways

Step-by-step explanation:

Part 1:

The number of ways in which we can organized n elements into k groups with size n1, n2,...nk is calculate as:

$\frac{ n!}{ n1!*n2!*...*nk! }$

So, in this case we can form 4 subgroups with 10 participants each one, replacing the values of:

• n by 40 participants
• k by 4 groups
• n1, n2, n3 and n4 by 10 participants of every subgroups

We get:

$\frac{ 40!}{10!*10!*10!*10!} = 4.7*10^{21}$

Part 2:

The number of ways in which we can choose k element for a group of n elements and the order in which they are chose matters is calculate with permutation as:

$nPk = \frac{ n!}{(n-k)!}$

So in this case there are 4 offices in the research board, those are director,  assistant director, quality control analyst and correspondent. Additionally this 4 offices are going to choose from a group of 5 doctors.

Therefore, replacing values of:

• n by 15 doctors
• k by 4 offices

We get:

$\frac{ 15!}{ (15-4)! } = 32760$

Answer 2

Answer:

847,660,528 combinations for part 1.

Step-by-step explanation:

Hi!  So I know this is a little over a year old, but I stumbled on this trying to find the answer for myself, so I figured I'd leave the correct answer here with an explanation, since all the others are bare-bones, at best.  

Basically, you have 40 volunteers and 10 slots to fill.  For the first slot, any of the 40 participants could be chosen.  For the second, however, there has to be someone that got picked for the space prior, so you only have 39 people to choose from and so on, until there are only 31 people left to pick by the time you get to slot #10.  the combination formula is read as : $n^{c} r$=$\frac{n!}{r!(n-r)!}$.  Here, n=40 because there are 40 options, and r=10, because there are 10 slots.  It would be difficult to do all the factorials of 40 (most calculators just refuse, which I found out the hard way), so this is where better understanding of the purpose of the formula becomes handy.  40-10=30, making 30!.  This is actually a representation of how many slots there are.  As I explained earlier, each space must decrease by one in number of possible options, which is why factorials are used in this situation, but it can't go all the way back to zero.  30! can cancel everything in 40! past 31, so there's only 40-31 left, the actual number of slots required.  10! will calculate, and it's the only thing left in the denominator.  This part of the process is necessary because the order doesn't matter in combination equations, so combinations where the only difference is the order cannot be counted. Divide from here, and you should be left with 847,660,528.

For part two, you must use the permutation formula:$n^{P} r$= $\frac{n!}{(n-r)!}$.  There are four slots and 15 doctors, and the order in which they are arranged after their selection does matter this time.  15-4=11, so $\frac{15!}{11!}$.  To divide, remove every repeating number, leaving only 15·14·13·12.  This should solve out to 32,760.

Please let me know if I missed anything or made any mistakes!