Math Problem
 

A music festival award gives awards for 1st, 2nd and 3rd place in each category. If there are 7 contestants for piano and 9 for violin, in how many different ways could the 6 awards be presented?

I don't understand the question. Wouldn't you give the awards to the 3 best contestants in each category? So, I guess my answer would be that there is only one way that is fair and makes sense.

There is 1st, 2nd and 3rd place for piano, and 1st, 2nd and 3rd place for violin. Well, the question also assumes that even the worst player could get 1st.

I believe what you are looking for are the number of permutations. If that is the case, the number is 210 for the first category and 504 for the second category.

LOL. So, the awards are randomly selected? I could solve it, but I have had too many adult beverages.

I think the best player should get first place. Why would you give the worst player an award? Not a math problem. Sounds like woke problem.

I took the problem from a school math test. The question is looking for the total possibilities…( and that’s where I made a mistake when I tried it. )

If you want the number of permutations of 6 winners (when considering only 1-2-3 place within each of the two categories), I believe the number is 105,840.

Yes! That’s where I made my mistake, I added instead of multiplied.(Standardized grade 7 math test.)

(7x6x5)x(9x8x7)

What Tuccillo said. :)

Interesting approach, just divide out the denominator.

For those who are curious, the general formula for permutations is:

n! / (n-s)!

where n is the number of objects you are considering and s is the number of objects in each set. Apply this to each of the two categories and then multiply.

First prize can go to any 1 of the 15 participants, So 15 choices.

2nd prize can then go to 1 out of the remaining 14.

And 3rd can go to any one of the remaining 13.

So number of ways we can do this is 15*14*13

3*5*2*7*13

3*10*91

2730 ways.

Awards Contest
 

For the piano (7 total participants): There are 210 possible arrangements (permutations) of 3 participants. 7 possibilities for 1st Place, 6 possibilities for 2nd place, and 5 possibilities for 3rd place. Multiplying these (7x6x5) gives you 210 possible arrangements.

For the Violin (9 total participants) : There are 504 possible arrangements (permutations) of 3 participants. 9 possibilities for 1st Place, 8 possibilities for 2nd place, and 7 possibilities for 3rd place. Multiplying these (9x8x7) gives you 504 possible arrangements.

So, the total number of ways the trophies can be presented is 7 x 6 x 5 x 9 x 8 x 7 = 105,840.

[ In other words the piano 1st place trophy would be given to one of seven participants; the piano 2nd place trophy would be given to one of the remaining six participants, and the piano 3rd place trophy would be given to one of the remaining 5 participants. Similarly, the violin 1st place trophy would be given to one of nine participants; the violin 2nd place trophy would be given to one of the remaining eight participants; and the violin 3rd place trophy would be given to one of the remaining 7 participants. ]

My grandson loves math.
I can't even understand any of the stuff he does.
Kids today are so far in front of where we were at same age.
I was a bit thick anyway, but I knew how to work out my hours and wages!

Simple Math
 

It’s a perm, because order does matter. Don’t treat it as a combination.

another vote for Tuccillo's answer

As an aside, my maternal grandmother's village in England held an annual Show and one year refused to give a 1st Place for the carrots as none of the entries was considered good enough. They just awarded 2nd and 3rd.

That's a total of 714 trophies, I should have been a trophy maker!

:popcorn: :popcorn:

When awards are handed out, often they award 3rd place first, then second, then first.

Actually, there are only 3 awards so THOSE 3 awards can only be present in 6 possible arrangements. That makes 6 and 6.

Most critical question, in how many different ways could the 6 awards be presented?
Each of the categories has 3 awards and can't be combined with the other so, if one is being realistic, catagory one has 3 awards and catagory two has three awards.

3 (of anything) only has 6 possible arrangements so the answer would be:
6 different ways for each category or 12 different ways for thoses 6 awards.

The performances were lousy. No awards given!

(7x6x5)+(6x5x4)= # of possibilities

The wording of the question could have been better. Regardless, I am pretty sure the intent was to compute the possible permutations, which was my answer. Otherwise, the number of participants in the two categories would not have been presented.

There are no 1rst, 2nd or 3rd place awards. Everybody gets a participation medal and complains that they had to practice too much.

There are so many ways this can be taken. Was permutation in the original question?

I hate when questions are not clear. In college I would clarify the various ways the question could be perceived, then solve for each. Drove the profs crazy.

I was on the board of a few schools in NY. I felt they were going the wrong way as far as teaching math went. In grade school, instead of providing the actual answer, they were being asked to estimate the answer. There is no estimate in math. There is one correct answer.

The act of holding the contest is triggering and a micro aggression. Shame on them.

————-—————42————————-

From Hitchhikers Guide to the Galaxy

So, if people who play piano don’t also play the violin so they can’t get an award for playing the violin, and people who play violin don’t also play the piano, would the answer be (9X8x7) + (7x6x5)?

567 combinations. (3 x 7 x 3 x 9)

Answer
 

There are a total of 6 awards, assuming per your problem statement the awards can be presented in any order, then the answer is 6! (6 factorial) or 720. This is the number of ways 6 distinct things can be orderd.

It's not clear from the question how many categories there are or how many awards are given in each category. In order to answer the question as stated, we would need more information. If there are two categories, piano and violin, and each category has three awards (1st, 2nd, and 3rd), then there are 7 * 3 = <<73=21>>21 different ways to award the piano prizes and 9 * 3 = <<93=27>>27 different ways to award the violin prizes, for a total of 21 * 27 = <<21*27=567>>567 ways to award all of the prizes.

There are two categories and 3 awards in each category. See posts #4, #7, and #10. This is, essentially, an exercise in computing permutations.