Question 1115996
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You have two responses from other tutors; both of them use the numbers 1 4 6 4 1 to find the answer.  One tutor refers to the numbers as a row of Pascal's Triangle; the other refers to them as a binomial expansion.<br>
And indeed the rows of Pascal's Triangle contain the coefficients of the binomial expansion of {{{(x+y)^n}}}.<br>
But let's take a look at WHY those are the numbers we want to use to solve this kind of problem.  In other words, let's see WHY binomial expansion works the way it does.<br>
If you understand why, the formula for binomial expansion will make sense; it will no longer seem like magic.<br>
I will guess that you are familiar with the FOIL method for multiplying two binomials.  The product of two binomials is the sum of four "partial products":
(1) F: the First terms in the two binomials,
(2) O: the Outer two terms,
(3) I: the Inner two terms, and
(4) L: the Last two terms.<br>
That familiar method of multiplying two binomials is a simple case of a general principle which says that the product of any number of polynomials is the sum of all the "partial products" obtained by choosing one term from each polynomial.<br>
In the expansion of {{{y+3)^4}}}, we are multiplying four identical factors of (y+3).  The y^3 term in the expansion is the sum of all the partial products in which the "y" term is chosen in 3 of the 4 factors and the "3" is chosen in the other factor.<br>
The number of ways of choosing 3 of the 4 factors from which to choose the "y" term is {{{C(4,3) = 4}}}.<br>
So the y^3 term is found by combining 4 partial products, in each of which the "y" term is used 3 times and the "3" is used once.  That combining gives us the complete y^3 term in the expansion:<br>
{{{(4)*((y)^3)*((3)^1)}}} = {{{12y^3}}}<br>
So the coefficient of the y^3 term in the expansion is 12.