How can i add rows of pascal's triangle without using an array?

[gelmi]

**Enter row number: 4
Sum of numbers in row 5 is 31. Sum of all numbers above 4 is 15.
this should be the output look like..
can someone help me? thanks.

[Nutster]

The elements of Pascal's triangle are the statistical combination of the given row and column, if you consider Pascal's triangle as a right triangle.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
...
Each element of Pascal's triangle is the result of nCr, where n is the row number, starting at 0, and r is the column number, also starting at 0. As far as I can tell, Combination is not a function in the Math library, but really it should be.

public static long Factorial(byte nValue)
{
    long nFact = 1L;

    // If nValue > 20, this will generate a numberic overflow, even using long.
    if (nValue < 0)
        return 0L;
    else if (nValue < 2)
        return 1L;

    for (long i=nValue; i > 1; --i)
        nFact *= i;
    return nFact;
}

public static int Combination(int nSetSize, int nChosen)
{
    int nLow;
    int nCombo = 1;

    if (nChosen > nSetSize-nChosen)
        nLow = nChosen;
    else
        nLow = nSetSize - nChosen;
    for (int i=nLow+1; i <= nSetSize; ++i)
        nCombo *= i;
    nCombo /= Factorial(nSetSize - nLow);
    return nCombo;
}

This code has not been tested, so there might be a bug or two in it.

There is also a good article on this stuff, including producing Pascal's triangle here.

[green_ning]

@gelmi

Are you in DOS mode?

[Nutster]

Actually the sum of any row a Pascal's Triangle is 2^n, where n is the row number, starting at 0. The sum of all values above a certain row is (2^n)-1. So the sum of all values in row 5 is 2^5 = 32. The sum of all values above row 5 is 2^5 - 1 = 32-1 = 31.