combinations and probability

Question

For some civil cases, at least 9 of 12 jurors must agree on a verdict. How many combinations of 9 jurors are possible on a 12 person jury?

Answer 1

First you want to check what kind of question this is. Based on the unit that you are working from, I think you know it's a permutations/combinations problem.

Next, check on whether it's a permutation or combination. For a permutation, remember "order is important," so the situation would be that we care about who agreed in what order. For a combination, we don't care about what order they agreed, just that they said guilty or not guilty. Clue words in problems that are usually used for combinations-

1. "combinations," "groups," "committees," or anything that shows that we don't care about what position, order, or title the people or things have.

Clue words for permutations-

1. "orders," "arrangements," sequences," or anything that tells us that it matters which came first, second, third, and so on.

Which one are we talking about here, permutation or combination?

For a permutation or combination you set up the number of items that we're choosing from as n and how many we pick as r. Here we are picking 9 jurors out of 12.

To get the answer, either do _nP_r or _nC_r depending on whether we have a permutation or combination here. You can plug numbers into a TI-84 by going to MATH -> PRB -> _nP_r or _nC_r.

If you are required to use the full equation, the _nP_r formula is n!/(n-r)! and the _nC_r formula is n!/((n-r)!(r!)).

Here's how I'd work through a very similar problem:

Packing for a 5am flight in the dark, Rebecca randomly grabs 3 shirts out of the 7 in her closet. How many different combinations of 3 shirts could she have?

1. Identify: order is not important so this is a combination problem.
2. Find n and r: the total number of shirts is 7 and the number she picked is 3, so n=7 and r=3.
3. Set up and solve: the problem is ₇C₃ (read aloud as "seven choose 3") so either use the calculator function or solve 7! divided by (7-3)!(3)! Remember that 7! means 7*6*5*4*3*2*1. This would end up being 7*6*5*4*3*2*1 divided by (4*3*2*1)(3*2*1). That is equal to 35.

Thanks for the question and I hope that this helps you with this unit, which is usually a challenging part of Algebra 2 for most students!

Answer 2

This is a combination since the order of selection does not matter.

_nC_r = n! / r!(n - r)!

₁₂C₉ = 12! / 9!(12 - 9)!

Can you calculate and answer?