For The Questions, a club has 10 members. a. Find the number of different slates of 3 officers (list of president, vice-president, and treasurer) that the club could have for officers. b.Find the number of different slates of 4 officers (list of president, vice-president, secretary, and treasurer) that the club could have for officers.

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For The Questions, a club has 10 members. a. Find the number of different slates of 3 officers (list of president, vice-president, and treasurer) that the club could have for officers. b.Find the number of different slates of 4 officers (list of president, vice-president, secretary, and treasurer) that the club could have for officers.

Mathematics
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Are you studying combinations and permutations?
Yes
Is this question a combination or a permutation? What do you think?

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Other answers:

Remember that order doesn't matter in a combination but order does matter in a permutation.
i think the first one is a permutation
Right you are. In fact, they both are. Can you see that?
Yes, because they have to be in a certain order, right?
So, in the first question, you need to calculate\[_3 P_{10}\]Can you do that?
I'm not sure how too
Actually, my notation was backwards. Sorry. The general equation for permutations is\[_{n}P_r=\frac{ n! }{\left( n-r \right) !}\]And in the first question n=10 and r=3. Can you take it from here?
I got -2 that's probably not correct is it?
Not quite. The equation becomes\[_{10}P_3=\frac{ 10! }{ 7! }\]If you write out the factorial multiplications you'll see that all the factors from 7 down will cancel.
Something like\[_{10}P_3=\frac{ 10\times9\times8\times7\times6\times5\times4\times3\times2\times1 }{ 7\times6\times5\times4\times3\times2\times1}\]
See. Everything from 7 down will cancel out leaving\[_{10}P_3=10\times9\times8=?\]
720? that's what I got from the multiplication
Excellent. Now try the second question by yourself. You need to calculate\[_{10}P_4\]
5040?
Terrific. Well done.
Thanks!
You're welcome

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