How to obtain the result of n(n-1)(n-2) / 6

Question

In my Python book, the question asks to prove the value of x after running the following code:

x = 0
for i in range(n):
    for j in range(i+1, n):
        for k in range(j+1, n):
            x += 1

What I could see is that:

i = 0;  j=1;  k=2:  from 2 to n, x+=1, (n-2) times 1
i = 1;  j=2;  k=3:  from 3 to n, x+=1, (n-3) times 1
...
i=n-3;  j=n-2; k=n-1: from n-1 to n, x+=1, just 1
i=n-2;  j=n-1; k=n doesn't add 1

So it seems that the x is the sum of series of (n-2) + (n-3) + … + 1?
I am not sure how to get to the answer of n(n-1)(n-2)/6.

Asked By: ACuriousCat

||

Source

Answer 1

Just write the for loops as a sigma: S = sum_{i=1}^n sum_{j=i+1}^n sum_{k = j + 1}^n (1).

Try to expand the sum from inner to outer:
S = sum_{i=1}^n sum_{j=i+1}^n (n - j) = sum_{i=1}^n n(n-i) - ((i+1) + (i+2) + ... + n) = sum_{i=1}^n n(n-i) - ( 1+2+...+n - (1+2+...+i)) = sum_{i=1}^n n(n-i) -(n(n+1)/2 - i(i+1)/2) = sum_{i=1}^n n(n+1)/2 + i(i+1)/2 - n*i = n^2(n+1)/2 + sum_{i=1}^n (i^2/2 + i/2 - n*i).
If open this sum and simplify it (it is straightforward) you will get S = n(n-1)(n-2)/6.

Answered By: OmG

Answer 2

One way to view this is that you have n values and three nested loops which are constructed to have non-overlapping ranges. Thus the number of iterations possible is equal to the number of ways to choose three unique values from n items, or n choose 3 = n!/(3!(n-3)!) = n(n-1)(n-2)/3*2*1 = n(n-1)(n-2)/6.

Answered By: pjs

Answer 3

This is a mathematical model of the algorithm, known as binomial coefficient (combinations). In analysis of algorithms you can use this form : N^k/k!
In your code you have a case without repetition that’s why n*(n -1 )*(n -2 ) / 3!

Answered By: Artur

How to obtain the result of n(n-1)(n-2) / 6

Question:

Answers: