Question 1096350: Hi, I am having trouble figuring out how to solve this following problem:
A real estate office has 10 sales agents. Each of four new customers must be assigned an agent.
(a) Find the number of agent arrangements where order is important.
(b) Find the number of agent arrangements where order is not important.
I have considered using methodologies such as assigning a customer name to each sales agent such as assigning customers 1,2,3,4 to agent 1 all the way to 37,38,39,40 to agent 10 and attempting to switch the customers around so that every possible combination is found. I know this is a combination problem, but I do not know how to apply the formula.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
The note you posted, where you mentioned "I have considered using methodologies such as assigning a customer name to each sales agent such as assigning customers 1,2,3,4 to agent 1 all the way to 37,38,39,40 to agent 10", is a bit confusing. Why are you counting all the way up to agent 40? There are only ten agents.
I think you should strike out the "to 37,38,39,40" like so
I have considered using methodologies such as assigning a customer name to each sales agent such as assigning customers 1,2,3,4 to agent 1 all the way to 37,38,39,40 to agent 10
If you do that, then it would make a lot more sense. Anyways, onto the problem.
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Part A)
Imagine we have four empty slots (much like a slot machine) where the agent numbers (1 through 10) can be assigned. Each slot represents a customer. Call these slots A, B, C, and D.
Slot A: we have 10 choices for this first slot as there are 10 agents to choose from.
Slot B: after the first customer (in slot A) chooses their agent, there are 10-1 = 9 agents left
Slot C: Similar to the previous slot: we have 9-1 = 8 choices left
Slot D: Similar to the previous slot: we have 8-1 = 7 choices left
Notice we have a countdown going. We start at 10 and stop after counting exactly four items down.
So we have the countdown: 10, 9, 8, 7.
Multiply out these four values:
10*9*8*7 = 90*56 = 5040
The value 5040 is the final answer to part A.
An alternative route is to use the nPr permutation formula. Permutations apply to determining how many ways we can arrange a group (or subgroup) of items. Order matters.
Here we have n = 10 as our overall sample size and r = 4 as the selection size
Plug those values into the nPr formula below to get...
nPr = (n!)/((n-r)!)
10P4 = (10!)/((10-4)!)
10P4 = (10!)/(6!)
10P4 = (10*9*8*7*6!)/(6!)
10P4 = 10*9*8*7 .... notice how the "6!" terms cancel
10P4 = 5040
and we get the same result as before (5040)
Side Note 1: The exclamation mark means factorial indicating we start at the given value and count our way down to 1. We multiply all of these values in the countdown (eg: 5! = 5*4*3*2*1 = 120). See the "further reading" section at the bottom of the page for more info about factorials.
Side Note 2: With expanding out 10!, we could do 10*9*8*7*6*5*4*3*2*1 which is a bit cumbersome. Luckily we can stop at 6 leaving the factorial symbol. That allows for easily cancellation. See the mini proof below for confirmation how this works.
Side Note 3: 10P4 = 5040 = 7! (A bit curious. I'm not sure if it's a coincidence or something stronger. I'll have to look into it later)
Proof for side note 2 above:
10! = 10*9*8*7*6*5*4*3*2*1
10! = 10*9*8*7*6*5*4*3*2*1
10! = 10*9*8*7*(6*5*4*3*2*1)
10! = 10*9*8*7*(6!)
10! = 10*9*8*7*6!
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Part B)
Unlike with part A, we now don't care what the order is. For example, the sequence {10,9,8,7} is identical to {7,8,9,10}. Order does not matter. For any group of four items, there are exactly 4! = 4*3*2*1 = 24 ways to arrange the group. This means that we have 24 times more listings than we want. So simply divide the result from part A (5040) by 24 to get: 5040/24 = 210
The answer to part B is 210.
Following the formula path, we can use the nCr combination formula. With combinations order does not matter.
Side Note: A typical locker is often secured with a "combination" lock, which is a misnomer. It should be called a permutation lock. If it were truly a combination lock (in the context of algebra), then a sequence like "1-2-3" is identical to "3-1-2". However, those sequences are clearly different for purposes of securing valuables.
Use the nCr formula, with n = 10 and r = 4 as before, to get...
n C r = (n!)/(r!(n-r)!)
10 C 4 = (10!)/(4!*(10-4)!)
10 C 4 = (10!)/(4!*6!)
10 C 4 = (10*9*8*7*6!)/(4!*6!)
10 C 4 = (10*9*8*7)/(4!) .... again the "6!" terms cancel
10 C 4 = (10*9*8*7)/(4*3*2*1)
10 C 4 = (5040)/(24)
10 C 4 = 210
We get 210 again. Notice on the second to last step we have (5040)/(24) which confirms the previous line of thinking. The connection between permutations and combinations is such that we simply divide the permutation result by r! (r factorial) to get the combination value we want. In other words
n C r = (n P r)/(r!)
10 C 4 = (10 P 4)/(4!)
10 C 4 = (5040)/(24)
10 C 4 = 210
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In summary:
The answer to part A is 5040
The answer to part B is 210
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Further Reading (optional but encouraged)
All of the articles are from the same website. They all connect and link together in some way. So (in hindsight) it probably only makes sense to just post one link. However I'll leave it as it is for you to explore.
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