combinations

Without repetitions there are P(88,5) = 88*87*86*85*84 possible sequences.

With one repetition there are C(88,4)*4 ways to pick the keys and 5!/2 ways to arrange them.

C(88,4) = 88*87*86*85/(4*3*2*1)

So you get P(88,5) + C(88,4)*4*5!/2 total,

which is P(88,5) + P(88,4)*10.

That works if the first key is the repeated one, but if you include the conbinations of doubled keys you multiply your answer by 5!/[1!(5-1)!]=5

Rap

i agree that there are P(88,5) (whats the name for this useage with the P with numbers?) that is 88*87*86*85*84 five note sequences on the piano without any repeats.

if we look at one repeat. well it would only use up 4 of the 88 keys so, P(88,4) is the initial amount of four note sequences available, ie 88*87*86*85, and for each of these 4 notes, repeating one of them and changing the position the note is repeated in, gives you 16 ways of arrangement, so that would be 88*87*86*85*16

= P(88,5) + P(88,4)*16
= 88*87*86*85*84 + 16*88*87*86*85
= 88*87*86*85*100

is this not true? i'm not quite sure what you did mark, perhaps you could help explain what i did wrong here

aradesh / Rap:

P(n,r) = n!/(n-r)! = the number of permutations of n things taken r at a time
C(n,r) = n!/[(n-r)!*r!] = the number of combinations of n things taken r at a time

First, let's choose the keys:
There are C(88,4) ways to choose four non-repeating keys. There are then four ways to choose a repeated key.

Now, let's arrange the five keys:
Five keys with one repetition can be arranged in 5!/(2!*1!*1!*1!) ways. In general, n objects with ri occurrences of the ith object can be arranged n!/(r1!*r2!*...*rm!) ways. The sum of the ri's is n. (In the preceding two sentences, the i's and the m were intended to be subscripts). For example, of you had allowed three of one key and two of another, the number of arrangements (once you selected the keys) would have been 5!/(3!*2!).

Therefore, there are C(88,4)*4*5!/2 ways to choose and arrange five keys with one repetition.

aradesh: I hope that explains it better.

Rap: I don't know why you're suggesting that this is off by a factor of five.

This is made up right?

It's a field called combinatorial analysis - I think it's always the first few weeks of every probability or statistics course. Lots of books in the book stores.

Combination of 4 at a time

C(88,4)=88!/(88-4)!=88*87*86*85

the 5th key being one of is the number of combinations is P(4,1)=4!/(4-1)!1=4

e.g.
AABCD
ABACD
ABCAD
ABCDA

so

C(88,4)P(4,1)=4*88*87*86*85 ((?))

Rap

Actually, that's P(88,4). C(88,4) = 88!/[(88-4)!*4!]

Once again:

Step 1: Select four keys
C(88,4) = 88*87*86*85/(4*3*2*1)

Step 2: Select which key to repeat
C(4,1) = 4 (I'm calling this C(4,1) instead of P(4,1) because we're not arranging the choice, just selecting it)

Step 3: Arrange the five keys taking into account that there is one repetition
5!/(2!*1!*1!*1!) = 5!/2!

Step 4: Multiply the results of steps 1-3 to obtain the final answer
C(88,4) * P(4,1) * 5!/2! = 88*87*86*85/(4*3*2*1) * 4 * 5!/2! = 88*87*86*85*10

OK

I see 1 and 2, but three gives me an itch

If I have five unique objects then the number of permutations is 5!/1!, if I have two that are identical then the number of permutations p(5,2) is 5!/2!. Where did those other 1! come from.

Nevertheless 5!/2!=5!/2!/1!/1!/1! so the number comes out as you described it cause

C(88,4)C(4,2)P(5,2)=[88*87*86*85/((88-4)!4!)]*[4!/((4-1)!1!)]*5!/2!=10*88*87*86*85

Rap

The extra 1!'s are there for completeness to indicate that the other three keys are not repeated. They are not necessary.

ok mark i managed to follow your way. i'll try explaining it and see if my explanation is a correct interpretation of the working out:

concerning the 5 notes where one is repeated:

for these, there are only 4 unique keys played so the number of four note chords (groups of notes in no particular order) on the keyboard is C(88,4).
for each four note chord, there are 4 ways of repeating a note (ie four possible notes which can repeat). for each of these instances there are 5!/2 ways of arranging these notes about.
giving total number of combinations of 10*P(88,4) as you showed. however, where is the flaw in my working which lead to a slightly larger answer? i presume it repeats some ways twice, but can't really see how exactly.

Can you show your formula?

here is what i put earlier

So life is about combinations and alot of these are both in numeric form and others in the juxtaposition of natural acts intermixed with other natural acts and thus evolving into reality.Now you take that reality and study it specifically in the realm of physical objects and eventually your math has to evolve into that of both concept and flexibility.Thus we start solving the riddles of this physical unverse using models as show to us above. Is that it ? Is that what your trying to get us to see ? That numbers ahve a life of there own regardless of outr needs and understanding of them ?

aradesh:

you seem to be multiplying the four repeated notes by four different arrangements. There are four ways of selecting a repeated note. There are then 5!/2! ways of arranging all five notes. Normally, there would be 5! ways to arrange five notes. However, since there is a repetition, in every arrangement, we can rearrange the repeated notes, so we divide by 2!.