When (x+y)n-1 is multiplied by x+y, each coefficient in the product, other than xn and yn, is the sum of two coefficients from the previous product. For example, every instance of x3y6 in (x+y)9 comes from x2y6×x or x3y5×y. Hence we add the coefficients on x2y6 and x3y5, from (x+y)8, to get the coefficient on x3y6 in (x+y)9. The coefficients on the terms xn and yn are always 1. This mirrors the construction of Pascal's triangle perfectly.
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There is another formula for the jth coefficient in the product.
By the distributive law, there are 2n terms,
selecting x or y at each step.
The number of terms with j instances of x and n-j instances of y
is equal to the number of
combinations of j elements, drawn from n elements.
This is n choose j, written (n:j), with formula
n!/(j!×(n-j)!).
This means the jth number
in the nth row of Pascal's triangle is also (n:j).
Since these numbers are the coefficients on the nth power of a binomial, they are often called binomial coefficients. Set x=y=1 and obtain the immediate corollary: 2n is the sum of (n:i) as i runs from 0 to n. The sum of all the binomial coefficients from 0 to n, or the sum of the entries in the nth row of Pascal's triangle, is 2n. When x and y are 1 and -1, the alternating sum of binomial coefficients is 0. |
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The binomial theorem generalizes to the multinomial theorem when the original expression has more than two variables, although there isn't a triangle of numbers to help us picture it. When x+y+z is raised to the n, there are n!/(i!×j!×k!) ways to arrange i instances of x, j instances of y, and k instances of z. Hence this combination becomes the coefficient on xiyjzk.
Let's look at (x+y+z)7. The coefficient on xy2z4 is the number of ways you can arrange xyyzzzz. That's 7! divided by (1!×2!×4!), or 105.