Combination Calculation. Please Help.

[Maths Noob]

Hi there,

This is my first post, I'm useless at maths, but I want to get better.

I have 3 separate columns containing 7 music clips each, each column is playing 1 of the 7 clips at random simultaneously.

How do I calculate how many combinations can be played?

Is it 7 x 7 x 7 =7³ ?

That gives me 343, but that seems too small.

Is there some kind of random-variable-permutation principle I'm missing?

What about factoring in repeats?

Please help, my brain hurts!

Thanks in advance

Regards

Maths Noob

 

[Dr.Peterson]

You aren't useless; you got it right! You just don't have the right intuition -- and it takes a lot of experience to get that.

If you want to convince yourself that the answer is correct, try listing all the possibilities.

I'm not sure what it means for a column to play music, but I don't need to know! If the first can be A, B, C, D, E, F, or G, and the second can be H, I, J, K, L, M, or N, and the third can be O, P, Q, R, S, T, or U, then start with AHO, then AHP, and so on to GNU, and you should find you've listed 343 "words". (Or you could do it with A, B, C, D, E, F, or G for each of them: AAA, AAB, ..., GGG.)

Or think about numbers. How many 3-digit numbers are there? Each digit has ten choices; and from 000 to 999 there are 1000 numbers. So your answer has to be less than that!

 

[wendellbernard]

Hi there,

This is my first post, I'm useless at maths, but I want to get better.

I have 3 separate columns containing 7 music clips each, each column is playing 1 of the 7 clips at random simultaneously.

How do I calculate how many combinations can be played?

Is it 7 x 7 x 7 =7³ ?

That gives me 343, but that seems too small.

Is there some kind of random-variable-permutation principle I'm missing?

What about factoring in repeats?

Please help, my brain hurts!

Thanks in advance

Regards wacky flip

Maths Noob

You’re actually correct! Since each of the 3 columns picks 1 of 7 clips independently and repeats are allowed, the total number of combinations is 7×7×7=343. It might seem small, but that’s the right count-like rolling three 7-sided dice.

 

[JoshuaSass]

Nice job on getting this! Took me a second to think.

If you want some help, here's what always helps me to make combination lists.
Say I have a 10 digit number pad, and the code is 4 digits. I completely forgot my code and I need to go through each and every code till I find it (will take FOREVER but its to explain). I want to find out how many codes there are that can be made. So, the options (Numbers 1-10) will always be put to the power of however many numbers I can pick, so in this case, 10^4!

If, say, I wrote down that the last 2 digits have to be within 1-5, and the first 2 are 0-9. Now, I can lower this down. In this case, I can use the same formula, but slightly changed!

For the first 2 digits, the equation is just 10^2, but for the last 2 digits, since there is only 5 numbers, we would do 5^2

So, to know how many pin combinations there are, if you know the last 2 digits have to be 1-5, you get 10^2 * 5^2

 

[scottbrandon]

Hi there,

This is my first post, I'm useless at maths, but I want to get better.

I have 3 separate columns containing 7 music clips each, each column is playing 1 of the 7 clips at random simultaneously.

How do I calculate how many combinations can be played?

Is it 7 x 7 x 7 =7³ ?

That gives me 343, but that seems too small.

Is there some kind of random-variable-permutation principle I'm missing?

What about factoring in repeats?

Please help, my brain hurts!

Thanks in advance

Regards

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Your math is correct, 343 is the right answer. Seven options in each of three independent columns is simply 7 x 7 x 7 = 343. Repeats are already factored in because each column picks independently regardless of what the others pick. It feels small but the number is right, your brain was working fine.