Given a natural number n, find the number of ways in which n can be expressed as a sum of natural numbers when order is taken into consideration. Two sequences that differ in the order of their terms define different compositions of their sum.
Input : 4 Output : 8 Explanation All 8 position composition are: 4, 1+3, 3+1, 2+2, 1+1+2, 1+2+1, 2+1+1 and 1+1+1+1 Input : 8 Output : 128
A Simple Solution is to generate all compositions and count them.
Using the concept of combinatorics, it can be proved that any natural number n will have 2^(n-1) distinct compositions when order is taken into consideration.
One way to see why the answer is 2^(n-1) directly is to write n as a sum of 1s:
n = 1 + 1 + 1 +…+ 1 (n times).
There are (n-1) plus signs between all 1s. For every plus sign we can choose to split ( by putting a bracket) at the point or not split. Therefore answer is 2^(n-1).
For example, n = 4
4 = 1 + 1 + 1 + 1 [We write as single 4]
Different ways to split once
4 = (1) + (1 + 1 + 1) [We write as 1 + 3]
4 = (1 + 1) + (1 + 1) [We write as 2 + 2]
4 = (1 + 1 + 1) + (1) [We write as 3 + 1]
Different ways to split twice
4 = (1) + (1 + 1) + (1) [We write as 1 + 2 + 1]
4 = (1 + 1) + (1) + (1) [We write as 2 + 1 + 1]
4 = (1) + (1) + (1 + 1) [We write as 1 + 1 + 2]
Different ways to split three times
4 = (1) + (1) + (1) + (1) [We write as 1 + 1 + 1 + 1]
Since there are (n-1) plus signs between the n 1s, there are 2^(n-1) ways of choosing where to split the sum, and hence 2^(n-1) possible sums .
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