5

In how many different ways can we select a committees consisting of 3 engineers, 2 biologists, and 2 chemists from 10 engineers, 5 biologists, and 6 chemists? First...

Question

In how many different ways can we select a committees consisting of 3 engineers, 2 biologists, and 2 chemists from 10 engineers, 5 biologists, and 6 chemists? First guess.

In how many different ways can we select a committees consisting of 3 engineers, 2 biologists, and 2 chemists from 10 engineers, 5 biologists, and 6 chemists? First guess.



Answers

A
college
needs
two
additional faculty
members:
a
chemist
and
a
statistician.
In
how many
ways
can
these
positions
be
filled
if
there
are
five applicants
for
the
chemistry
position
and
three
applicants
for
the
statistics
position?

This problem is concerning a real life application of deciding whether to use a combination or a permutation. So in this problem, it asks How many ways can a committee of three freshmen and forge in years before him from eight freshmen and 11 juniors? So we have to understand the difference between a combination and a permutation permutations formula is an factorial over and minus r factorial. While combination is n factorial over R factorial times and minus r factorial, the combination formula has that extra term because it illuminates cases where the terms are repeated. Because, um, it doesn't matter. But the order is for a combination. However, in a permutation order matters in this situation. It doesn't seem the order matters because this just the groups of fresh mint engineers are being formed from indistinct freshmen and juniors. They're not being ranked in order, so we can use a combination to find out how many groups are. So to find how many times how many different combinations of freshmen you would do a combination. Eight. Choose three and do juniors. You would do combination 11 Hughes four, and these two combinations will give you a group of three and four Select, respectively. So you want to know how many different ways they line up against each other to find these full seven people councils? So in that case, you talked to multiply these two combinations by each other to get the final answer. So 8 to 3 ends up being a factorial over bye factorial over three factorial times 11 factorial over seven factorial over four factorial and these go to end up being eight. Choose three is 56 times 11 who's four is 330 and that ends up being 18,480 and that is the final answer.

Okay, so we're finding the number of ways that three freshmen and four junior's could be chosen for a committee. And when I see the word committee, I don't see that it really matters what order the people are in, because if you're on a committee, you don't necessarily have a particular role. It doesn't say there's a president, vice president, et cetera, so I would think we would want to do this problem with combinations. And so we have three freshmen and four juniors and because of the and we're gonna use the multiplication principle and multiply our two combinations together. For the freshmen, the combination is eight is in, and three is our. And for the juniors, a combination is combinations of 11 and 4 11 juniors, and they're choosing forthem. If you multiply those together, you get 18,480. So that's my interpretation of the problem. Now the answer given in the book uses permutations so that either means that someone wanted committees to, um imply that order matters or someone made a mistake in finding the answer in the book. So if order matters and I think it shouldn't for the wording of this problem. If order matters, you're gonna do permutations instead. So be good to get some clarification from your instructor on that. Oops, This one's also permutations. That's P and that will give you let's see, 2,661,120.

So this question is about to select straight women from my transfer from a 25 members. So how many different commentators can be selected now? We can see we don't care. The older off thus remembers. So it's a combination problem. So we use see 25 three to represent the young sir. And we can see because that does represent, uh, combination. And that seacoast to the factory on 25 over the factorial off three. But because it's a combination and 25 minus three that is 22 and thus equals two, 2000 and 300. So we got through 2300 ways to select number.

Okay. This question has a situation where we have 10 people. And from there we're signing four of them, two unique committees. So the key thing here is that they're Scient, two unique committees. So it matters who goes to which one. So that should tell us that we're dealing with a permutation here. So let's say we have the four different committees right here. So for the 1st 1 we have 10 choices of who we could seat there. Then for the second, we have nine choices than for third. We have eight in the fourth. We have seven. And that gives us our total number, which is the same thing as the number of permutations from a group of 10. Before signing to four different committees and punching this into our calculator. Here we get 5040 different committee orders.


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