Question

Can you please check my answer?

You own 9 hats and are taking 2 on vacation. In how many ways can you choose 2 hats from the 9?

Answer: 362880

Since these appear to be permutations, I simply did the math as it follows:

9*8*7*6*5*4*3*2 = 362880

Answer 1

Wait a moment. If you taking one hat, how many ways can you choose one?

Answer 2

1?

Answer 3

No, there are 9 possibilities for hats. There are 9 ways to choose 1 hat out of 9.

Now, how many ways can I choose a first hat and a second hat? Well I can choose a first hat 9 ways, how many ways can I choose a second hat?

Answer 4

Does order matter? It should not be

9 Permutation 2

but

9 Combination 2

They are both in the suitcase ....

Answer 5

No, order doesn't matter.

Answer 6

No, it doesn't, but I want to get you to the answer. You can choose a first hat 9 ways; how many ways can you choose a second hat?

Answer 7

There are 8 hats left, so 8 ways to choose a second hat. So there are
9 x 8 ways
to choose a first and second hat. Agreed?

Answer 8

I just figured out the formula.

Answer 9

Is your answer smaller than 72?

Answer 10

no nvm

Answer 11

No.

Just to finish the logic we've developed, there are 9x8 = 72 ways to choose a first and second hat.
But now we don't care which is first or second; we can swap them. So we divide that number by 2

72/2 = 36.

There are 36 ways to choose two hats from a collection of 9.

Answer 12

The answer is a number smaller than 72, it is not the permutation formula, it is a combination. JamesJ will help you ...

Answer 13

Make sense?

Answer 14

Yes. Thank you.

Answer 15

So, if you're given:

2 bananas are to be selected from a group of 7. In how many ways can this be done?

Answer 16

Use the same logic, walk me through it.

Answer 17

7 * 6 = 42

Answer 18

42/2?

Answer 19

yes

Answer 20

Thank you, sir. I appreciate it.

Answer 21

I think the general formula for this type of problem would be:
\[n!/r!(n-r!)\]
where r = number to be chosen, and n = number to be chosen from (siz of the group)

Answer 22

There's a generalization of all this that someone might give you as an answer here. The more important thing is you understand how to calculate these things from first principles. If you understand that, then the generalization will make complete sense to you as well.

Answer 23

Yeah. You are correct.

Answer 24

you do the
\[ n! \] to find out all possible ways of arranging the whole group then divide by \[ r! \] so that you get the possible combinations of how many you want to choose (where order is taken into account) and then \[ (n - r)!\] so that order of the choices isnt taken into account