How many melodies? #2

There are $4$ ways to choose the last note (we cannot use E,F and G). Next we can choose the second note. There are $6$ ways of doing that, because we can't choose the one we used as the last one. Using the same reasoning we can choose the first note from $5$ remaining unused notes. In total this gives $4*5*6=120$ melodies.

Let T be the total number of distinct possibilities of choosing the keys, then T = 7 * 6 * 5 = 210. But we know that T includes the possibilities such that the arrangement ends with E, F, G, so we need to substract them, x1, x2 and x3 respectively.

We know that x1 = 6 * 5 * 1 because we are forcing to end it with E.

x1 = x2 = x3.

So answer = T - 3 * x1 = 210 - 90 = 120.

At first, let's consider the impossibilities. If we have to end the last note with E,F or G, there are 3 ways to do it. Chosing the end note, Now we are left with 2 positions to be filled with 6 notes in hand which can be done in 6P2 or 6 x 5 = 30 ways. Since we can choose the last note in 3 ways and the first two notes in 30 ways, there can be 3 x 30 =90 ways to create a melody which will end with E, F or G.

But the total ways of creating a melody without the end note condition is to choose 3 positions with any 7 notes which can be done in 7 x 6 x 5= 210 ways. Now if we eliminate the impossible, whatever remains, however improbable must be truth. So, the number of melodies that can be played without choosing the last note as E, F or G has to be = 210-90 = 120.

There are other elegant solutions. Just wanted to share mine. :)

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We have to choose three from the set "ABCDEFG".

The last note cannot be "EFG" . Make a partition of the given set like this :- "ABCD | EFG".

As the first two notes can take from the second part , i.e "EFG" , we get we get the following cases :

• Case 1 :- 0 Notes taken from second part (i.e. "EFG) by the first two notes which means we have to choose the all three notes from "ABCD" , we can do it in 4 X 3 X 2 ways . That is 24 ways

• Case 2 :- 1 Note taken from second part by the first two notes. That is the first note can chose from "EFG" in 3 ways and the second note from "ABCD" in 4 ways and the third note must be chosen from first part "ABCD" of which 1 was already selected in second note. Hence 3 ways for third note. Therefore we have is 3 X 4 X 3 i.e. 36 ways . We should also count the possibilities where the first note is chosen from "ABCD" and second from "EFG", which will again have 36 ways. Hence the total number of ways is 36 + 36 which is 72 ways.

• Case 3 :- 2 Notes taken from second part by the first two notes. That is both, the first and second note take from "EFG". The number of possibilities are 3 X 2 and third note takes from "ABCD" that is there are 4 ways. The total number of ways are 3 X 2 X 4 that is 24 ways.

Now adding all the cases , ( we are using sum rule here as we have three ways Case 1 or Case 2 or Case 3 to choose from) :- The total number of ways are :- 24 + 72 +24 = 120 ways.

Begin by determining total number of possible 3 note combinations. Since there are 7 notes, the permutations are 7 6 5 or 210. Since there is an equal distribution, each note may be used up to 30 times in various unique combinations. Since 3 notes are not permitted in the final position, 210-(3*30)=120.

$4(3\times2+3\times3)+3(4\times3+2\times4)=\boxed{120}$

Sorry, I cannot explain as I don't know how to convert math equations to latex in these sites