An inline skate has 4 wheels. How many ways could 4 replacement wheels be chosen from a pack of 10 wheels and fitted to a skate?

No need to think of removing Chris's points!  There are 10 possible choices for the first wheel.  For each of these 10 there are 9 possible choices for the next wheel.  For each of these 10*9 there are 8 possible choices for the next wheel.  For each of these 10*9*8 there are 7 possible choices for the last wheel.  

$${\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{7}} = {\mathtt{5\,040}}$$

From the package, we can choose any 4 of the 10. Thus, we have C(10,4) =210 ways of choosing the wheels.

But this only answers part of the question. On the skates themselves, we have 4 ways to choose which wheel goes on first, 3 ways of choosing the next one to go on, 2 ways for the next, and 1 way for the last.

So, the total number of ways to fit the wheels to a skate are 4 x 3 x 2 x 1  = 24.

So, the total ways to select the wheels from the package, times the number of ways to fit them to a skate = 210 x 24 = 5040.

I think this is correct, but I'd like someone to "check" me!!

Hi Chris,

Probability is not my strong suit.  I didn't even think about the second part of your answer until you pointed it out.

Your answer sounds right to me! 

I'll give you the points and then if one of the others tells us your answer is wrong i will take the points away again.  How's that!

My theory is......points awarded.....stay FOREVER !!!

What even if it is wrong!!

It's like cards....once you lay them on the table....they're considered "played".....no "do-overs"

No I do not accept that!   

Thanks Alan,

I didn't think Chris's was wrong.  Your's is neater though!   

Chris might think i can't have his points but I've still got his sunnies!!!   And, he can't have them back.