Probability: Evaluate 2 choices from a list of 7 songs.

[orfhlaith]

Hi,

I do not undertstand this question given in class, I was able to complete all the other ones,but not this one.

Evaluate 2 choices from a list of 7 songs.

I was thinking was it 2!/7! but I do not think that is the correct answer.
Thanks

 

[MarkFL]

I am assuming you are being asked to compute how many ways there are to choose 2 songs from a list of 7. We'll call this number $\displaystyle N$. For the first song, we have 7 choices and for the second song, we have 6 choices. If the order of the two songs matters, then we take the permutations, i.e:

$\displaystyle \displaystyle N=7\cdot6=\frac{7!}{5!}\equiv\,_7P_2$

However, if order doesn't matter, then we have:

$\displaystyle \displaystyle N=\frac{7\cdot6}{2}=\frac{7!}{5!(7-5)!}\equiv\,_7C_2$

 

[HallsofIvy]

I am not clear about the word "evaluate" here. I think that you mean, "how many different ways can you choose two items from a collection of 7 items?"
2!/7! is surely not the correct answer because it is less than 1!

There are 7 different items you could chose first. There are then 6 items left so there are 6 ways to choose the next. So there are 7(6)= 42 ways to do that. That can also be written as $\displaystyle \frac{7!}{5!}$.

But that is choosing a specific item first then another item. If the order is not important, then we have to divide by 2 because we don't want to distinguish between choosing item 5 and then item 4 and choosing item 4 and then item 5. In the case that we do not want to distinguish order, then the number of ways to do that is $\displaystyle \frac{7!}{2(5!)}= \frac{7!}{2!5!}$ which is the "binomial coefficient" also known as "7 choose 2".