<h3>Answers:</h3>
- Problem 10) There are 220 combinations
- Problem 11) There are 126 combinations
- Problem 12) There are 154,440 permutations
- Problem 13) There are 300 different ways
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Explanations:
Problem 10
The order of the toppings doesn't matter. All that matter is the group itself. We'll use the combination formula nCr = (n!)/(r!*(n-r)!) where n = 12 and r = 3 in this case.
So,
nCr = (n!)/(r!*(n-r)!)
12C3 = (12!)/(3!*(12-3)!)
12C3 = (12!)/(3!*9!)
12C3 = (12*11*10*9!)/(3!*9!)
12C3 = (12*11*10)/(3*2*1)
12C3 = 1320/6
12C3 = 220
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Problem 11
Like with problem 10, the order doesn't matter. This is assuming that each member on any given team has the same rank as any other member.
If you used the nCr combination formula, with n = 9 and r = 5, you should get the answer 126
Here's another way to get that answer.
There are 9*8*7*6*5 = 15120 different permutations. If order mattered, then we'd go for this value instead of 126
Within any group of five people, there are 5! = 120 different ways to arrange them. So we must divide that 15120 figure by 120 to get the correct value of 126 combinations
15120/120 = 126
Note the connection between nCr and nPr, namely,
nCr = (nPr)/(r!)
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Problem 12
Now this is where order matters, because the positions in basketball are different (eg: a point guard differs from a center).
We have 13 choices for the first position, 12 for the second, and so on until we reach 13-r+1 = 13-5+1 = 9 as the number of choices for that last slot.
So we'll have 13*12*11*10*9 = 154,440 different permutations
Now if the condition that "each player can play any position" isn't the case, then the answer would very likely be different. This is because for the center position, for instance, we wouldn't have 13 choices but rather however many choices we have at center. To make the problem simpler however, your teacher is stating that any player can play at any slot. Realistically, the answer would be far less than 154,440
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Problem 13
We have 6 applications for the 2 math positions. Order doesn't matter. That means we'll have 6C2 = 15 different ways to pick the math people. Use the nCr formula mentioned in problem 10. Since we'll use this value later, let's make x = 15.
There are 2 people applying for the chemistry teaching position, meaning there are 2 ways to fill this slot. We could compute 2C1 = 2, but that's a bit overkill in my opinion. Let y = 2 so we can use it later.
Similarly, there are 10 applicants for the Spanish teacher position, leading to 10 ways to get this position filled. You could compute 10C1 = 10 if you wanted to. Let z = 10 so we can use it later.
Once we figured out those x,y,z values, we multiply them together to get our final answer: x*y*z = 15*2*10 = 30*10 = 300
There are 300 different ways to select 2 math teachers, a chemistry teacher, and a Spanish teacher from a pool of 6 math applicants, 2 chemistry applicants, and 10 Spanish teacher applicants.
