Basic algebra

I found a new interpretation (atleast new to me) for (a+b)ⁿ ..if we have n items and we want to partition them into two parts, then (a+b)ⁿ gives us a clue..
(a+b)ⁿ can be interpreted as follows.
Imagine that you have n slots and each slot can take either a or b.
Let us come up with the number of ways in which these slots can be filled :
Assume n=3 for simplicity
In this case we have the following options
aaa
aab
aba
abb
baa
bab
bba
bbb

Now let us group this based on number  of a values and number of b values
aaa 

aab
baa
aba

abb
bab
bba

bbb

So we notice the following
a³b⁰ is occuring 1 time
a²b¹ is occurring 3 times
a¹b² is occurring 3 times
a⁰b³ is occuring 1 time

(a+b)³= 1xa³b⁰+3xa²b¹+3xa¹b²+1xa⁰b³

the number of times any combination aⁱb^j occurs is equal to the number of ways in which we can choose a i times and b j times while ensuring that number of a values and number of b values sum to n i.e. i+j=n

Apply the well known combination formula ⁿC_r = n!/(n-r)!r!
(a+b)³= ³C₀xa³b⁰+³C₁xa²b¹+³C₂xa¹b²+³C₃xa⁰b³


So (a+b)ⁿ = Σ _{r=0 to n} ⁿC_r x a^rb^(n-r)
Note that the coefficients are symmetric in the sense that a=r and b=n-r has the same coefficient as b=r and a=n-r
Suppose we set a=1 and b=1 then we get

2ⁿ =  Σ _{i=0 to n} ⁿC<sub>r


       OR  we can write this as</sub> 


2ⁿ=1+ⁿC₁+ⁿC₂+ⁿC₃+..ⁿC<sub>n


    OR we can write this as</sub> 


2ⁿ = 1+ n+n(n-1)/2 + n(n-1)(n-2)/6 + n(n-1)(n-2)(n-3)/24.... 


Let us now see how we can do the same thing for
(a+b+c)ⁿ


here we have to find the number of ways in which 3 kinds of items are selected
example we have apples, bananas and cherries
We need to select 3 fruit - how many ways can we do it
3 apples and 0 bananas and 0 cherries
2 apples and 0 bananas and 1 cherry
2 apples and 1 banana and 0 cherries
1 apple and 0 bananas and 2 cherries
1 apple and 1 banana and 1 cherry
1 apple and 2 bananas and 0 cherries
0 apples and 0 bananas and 3 cherries
0 apples and 1 banana and 2 cherries
0 apples and 2 bananas and 1 cherry
0 apples and 3 bananas and 0 cherries

This is like choosing r apples and s bananas and (n-r-s) cherries
Replace apples, bananas and cherries with a, b and c and we get the various terms in the equation
r can take value 0 to 3
s can take value 0 to 3
Replace 3 with n and we can get the solution for (a+b+c)ⁿ
So it is a double summation
(a+b+c)ⁿ = Σ _{r=0 to n} Σ _{s=0 to n} ⁿC_r,s x a^rb^sb^(n-r-s)

Now we can apply this same logic to find (a+b+c..+k)ⁿ where instead of choosing only a and b and c we
have to choose k items
This means we need to partition into k sets with set 1 having n₁ items, set 2 having n₂ items..and so on till set k-1 having n_k-1 items and the remaining items in set k
In this case each term in the summation is of the following form
aⁿ¹bⁿ²cⁿ³....j^njk^{(n-(n1+n2...nj))} and the coeffecient is n!/(n1! x n2! x n3!...(n-(n1+n2...+nj))!


Another interesting derivation

Suppose a=-1 and b = 1 and we substitute this in
(a+b)ⁿ  then we get the following results
when n is even
0 = 1 - n + n(n-1)/2 - n(n-1)(n-2)/6 + n(n-1)(n-2)(n-3)/24.... -n(n-1)/2 + n - 1
means
 n - n(n-1)/2 + n(n-1)(n-2)/6 -  n(n-1)(n-2)(n-3)/24.... =  1

when n is odd
0 = -1 + n - n(n-1)/2 + n(n-1)(n-2)/6 - n(n-1)(n-2)(n-3)/24.... 
means
n - n(n-1)/2 + n(n-1)(n-2)/6 - n(n-1)(n-2)(n-3)/24....  = 1

Each term will appear once with a positive sign and once with a negative sign. As a result it sums to zero.