There are 52 white keys on a piano. The lowest key is A. The Keys are designated A. B, C, D, E, F, G in succession, and then the sequence of letters repeats, ending with a C for the highest key.

A) if five notes are played together simultaneously, in how many ways could the notes al be 1) A's. 2) G's. 3) all the same letter. 4) different letters
B) if five keys are played in order, how would your answers in part a)'change?

A) Well, I must say, the piano must really love the letter A if it wants to play all the keys together in the chord! So, let's count the ways:

1) If all the notes are A's, you only have one option because the piano doesn't want to get too repetitive, it's not a parrot after all. So, there's just 1 way to play all A's together.

2) Now, if the keys were all G's, it would sound like a group of very enthusiastic frogs, but we still need to count the possibilities. Since there are 52 white keys and G happens to be the seventh note in the sequence A, B, C, D, E, F, G, we have 7 G's to choose from. So, we have 7 ways to play all G's together.

3) When you say "all the same letter," you have the freedom to choose any of the 7 letters without feeling guilty about leaving the others out. So, there are 7 ways to play the same note for all the keys.

4) Now, if we want to get creative and play different letters, it's like having a musical fridge with lots of unique magnets to choose from! We have all 7 letters available for each key, so it's a matter of multiplying the possibilities. That gives us 7 x 7 x 7 x 7 x 7 = 7^5 = 16,807 ways to play different letters simultaneously.

B) Ah, now we're playing with order! Let's see how things change:

1) If we're playing the keys in order, we need to start with A, and then we have 51 keys left to choose from for the next note. After that, we have 50 keys left, and so on. In this case, we have 51 x 50 x 49 x 48 x 47 = 543,744,000 ways to play the keys in order, all as A's.

2) Similarly, if we're playing all G's in order, we have 6 keys left to choose from for the next note, then 5, 4, 3, and finally just 2 keys left for the last note. So, there are 6 x 5 x 4 x 3 x 2 = 720 ways to play the keys in order, all as G's.

3) The concept doesn't change when we want them to be all the same letter in order. For each note, we need one of the 7 letters, so we have 7 x 7 x 7 x 7 x 7 = 7^5 = 16,807 ways to play them all as the same letter, but this time in order.

4) When we want to play different letters in order, each key has 7 options, just like before. So, the calculation remains the same: 7 x 7 x 7 x 7 x 7 = 7^5 = 16,807 ways to play different letters in order.

Now, let's get those fingers tickling those ivories and have some musical fun!

A)

1) If all five notes are A's, there is only one way to play them.
2) If all five notes are G's, there is also only one way to play them.
3) If all five notes are the same letter, there are 7 possible letters to choose from (A, B, C, D, E, F, G). Therefore, there are 7 ways to play them all the same.
4) If all five notes are different letters, there are 7 choices for the first note, 6 choices for the second note, 5 choices for the third note, 4 choices for the fourth note, and 3 choices for the fifth note. Therefore, there are 7 * 6 * 5 * 4 * 3 = 5,040 ways to play them with different letters.

B) If five keys are played in order, the number of ways to play the notes will be the same as in part A) for all cases except for case 3) (all the same letter). In this case, there will be only 5 ways to play them in order, as there are 5 different choices of letters for each note.

To answer these questions, we need to understand the concept of permutations and combinations.

Permutations: In permutations, the order of the elements matters. For example, "ABC" and "BAC" are different permutations.
Combinations: In combinations, the order of the elements does not matter. For example, "ABC" and "BAC" would be considered the same combination.

A) If five notes are played together simultaneously, we need to find the number of ways these notes can be played under different conditions:

1) If all the notes are A's, there is only one way to play them since they have the same letter.

2) If all the notes are G's, there is also only one way to play them since they have the same letter.

3) If all the notes are the same letter, we need to select a letter from A-G. Since only one letter is chosen, there are 7 options (A, B, C, D, E, F, or G) to choose from.

4) If all the notes are different letters, we need to select five different letters from A-G. To calculate this, we can use the concept of combinations. The number of combinations of selecting r elements out of n elements is given by the formula C(n, r) = n! / (r!(n-r)!). In this case, we need to select 5 out of 7 letters, so the number of combinations would be C(7, 5) = 7! / (5!(7-5)!) = 21.

B) If five keys are played in order, the answers from part A would change:

1) If all the keys are A's, the order in which they are played matters. Therefore, we need to calculate the number of permutations. Since there is only one letter, there is only one way to arrange them.

2) If all the keys are G's, again, the order in which they are played matters. So, there is only one way to arrange them.

3) If all the keys are the same letter, the order still matters. As there are 7 options to choose from (A, B, C, D, E, F, or G), the number of permutations for each letter would be 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040.

4) If all the keys are different letters, we need to select 5 different letters from A-G and find the number of permutations for each combination. Using the formula for permutations, the number of permutations of selecting 5 elements from 7 would be P(7, 5) = 7! / (7-5)! = 7! / 2! = 7 * 6 * 5 * 4 * 3 = 2,520.

Therefore, the answers in part A would change when considering the order in which the keys are played.

There are 7 octaves, plus an extra A,B,C

A)
Since there are 8 A keys, there are 8C5=56 ways to pick them.
For the G keys, there are only 7C5 = 21 choices.
For any key, consider it like this. Pick a key from the 1st octave. Now there are fewer octaves to pick the other 4 keys. 3 keys (ABC) will have 7 other octaves, and 4(DEFG) will have 6 other octaves. So, you can pick them in

3*7C4 + 4*6C4 = 165

For different letters, I expect you can figure that out.

B
If these were all permutations, you'd use 7P5 instead of 7C5, and so on.