Is this problem a combination or permutation?

Suppose a group of 10 must pick an assistant, a representative, a treasurer, and a recorder. How many ways can this be done?

Answer: Permutation Problem

There are 5040 permutations possible

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Explanation:

Order matters because the positions are different.

(A,B,C,D) is different from (B,A,C,D) because the first means person A is the assistant while the second means person B is the assistant.

We have 10 choices for the first slot, 9 for the second, 8 for the third, and 7 for the fourth. We count down the values because a person cannot be reselected to serve multiple positions simultaneously.

Multiply out those values to get 10*9*8*7 = 5040

There are 5040 permutations possible.

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As an alternative, you can use the nPr permutation formula with n = 10 and r = 4 like so

[tex]_n P_r = \frac{n!}{(n-r)!}\\\\_{10} P_{4} = \frac{10!}{(10-4)!}\\\\_{10} P_{4} = \frac{10!}{6!}\\\\_{10} P_{4} = \frac{10*9*8*7*6!}{6!}\\\\_{10} P_{4} = 10*9*8*7 \ \ \text{ ... note how this shows up again}\\\\_{10} P_{4} = 5040\\\\[/tex]

We get the same answer as before.

Answer:

Combination

Step-by-step explanation:

This is the answer because:

1) The difference between combinations and permutations is the ordering of each term.

Combinations - the order doesn't matter

Permutations - the order matters.

An example of this is, if you enter the number combination 5324 into your locker/cabinet, it won't open up because it is a different ordering and needs a specific combination which means it's permutation.  

2) In this situation, the order doesn't not matter since they just want different ways to select an assistant, a representative, a treasurer, and a recorder.

Hope this helps! :D

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