Pascal’s Triangle is not only an interesting mathematical work because of its hidden patterns, but it is also interesting because of its wide expanse of applications to other areas of mathematics. As one familiar with algebra may notice, the numbers in each row of the triangle are precisely the same numbers that are the coefficients of binomial expansions. For example, when one expands the binomial, (x + y)3, algebraically it equals 1x3 + 3x2y + 3xy2 + 1y3. The coefficients of this binomial expansion, 1 3 3 1, correspond exactly to the numbers in the third row of Pascal’s Triangle. In general, the nth row in Pascal’s Triangle gives the coefficients of (x + y)n.
The coefficients of Pascal’s Triangle can also be used in probability to find out how many subsets of r elements can be formed from a set with n distinct elements. The symbol for this is nCr, which is calculated by the following formula:
nCr = n! / [r!(n – r)!], where n! = n(n – 1)(n – 2) . . . (3)(2)(1).
In this formula, the n is equal to the row number of the triangle, while the r is equal to the element number in that particular row. For example, the number in the seventh row and third element place is 35. 7C3 should, thus, be equal to 35. Using the formula:
7C3 = 7! / [3!(7 – 3)!] = (7*6*5*4*3*2*1) / (3*2*1*4*3*2*1) = (7*6*5) / (3*2*1) = 35.
Consequently, if one wishes to find the number of subsets of 3 elements that can be formed from a set with 7 distinct elements, they would just have to look up the number at the seventh row and third element place in Pascal’s Triangle. Therefore, Pascal’s Triangle is a useful tool in finding, without tedious computations, the number of subsets of r elements that can be formed from a set with n distinct elements.
Its known applications in mathematics also extend to calculus, trigonometry, plane geometry, and solid geometry.