Coefficients Party, Part 2

the expansion of (a+b+c+b)^n will have terms of form (a^i)(b^j)(c^k)(d^l) where i+j+k+l = n

this is equivalent to diving n objects in 4 groups which can be done in (n+4-1)!/(n! * (4-1)!)

(X1+X2+.......+Xm)^n number of term = (n+m-1)C(n) for given question m = n = 4; 7C4 = 35

It is a very long problem

The following calculator can be used to simplify ANY polynomial expression.

Result (x+2a-5g+3)^4 = x^4+8ax^3-20gx^3+12x^3+24a^2x^2-120agx^2+72ax^2+150g^2x^2-180gx^2+54x^2+32a^3x-240a^2gx+144a^2x+600ag^2x-720agx+216ax-500g^3x+900g^2x-540gx+108x+16a^4-160a^3g+96a^3+600a^2g^2-720a^2g+216a^2-1000ag^3+1800ag^2-1080ag+216a+625g^4-1500g^3+1350g^2-540g+81

And by counting the number of terms the answer is 35.

Since we are looking for the number of terms, coefficients aren't important and thus it is far easier to simply ( x + y + z + 1 )^4 than ( x + 2a - 5g + 3 )^4, and both methods give us 35 terms.

It should be easy to see that the first term in the expansion ( x + y + z + 1 )^4 would be x^4, and the last would be 1. x^4 can be expressed as (4,0,0), where x = 4th power, y = 0th power, and z = 0th power. Likewise, 1 can be expressed as (0,0,0), where x = 0th power, y = 0th power, and z = 0th power.

Thus, you could find the number of ways 3 whole numbers can add up to 0, 1, 2, 3, 4. There's only 1 way to add up to 0 : (0,0,0). There's 3 ways to add up to 1 : (1,0,0) , (0,1,0) , (0,0,1). There's 6 ways to add up to 2 : (2,0,0) , (1,1,0) , (1,0,1) , (0,2,0) , (0,1,1) , (0,0,2). There's 10 ways to add up to 3 : (3,0,0) , (2,1,0) , (2,0,1) , (1,2,0) , (1,1,1) , (1,0,2) , (0,3,0) , (0,1,2) , (0,2,1) , (0,0,3). There's 15 ways to add up to 4 : (4,0,0) , (3,1,0), (3,0,1) , (2,2,0) , (2,1,3) , (2,0,2) , (1,3,0), (1,2,1) , (1,1,2) , (1,0,3) , (0,4,0) , (0,3,1) , (0,2,2) , (0,1,3) , (0,0,4).

This is just the sum of the first five triangular numbers, and 1 + 3 + 6 + 10 + 15 = 35

using the binomial theorem for ((x+2y) + (-5z+3))^4:

(x+2y)^4 = 5 terms

4((x+2y)^3)*(-5z+3)) = 8 terms

6(x+2y)^2*((-5z+3)^2) = 9 terms

4(x+2y)(-5z+3))^3 = 8 terms

(-5z+3)^4 = 5terms

This problem can be solved can be solved by diving the exponents which will make it easier to solve:

$(x+2y-5z+3)^{4} = ((x+2y+3-5z)^{2})^{2}$

If we solve it by applying the formula, we will get the expanded form as:

$(x+2y-5z+3)^{4}) = x^{4}+8x^{3}y-20x^{3}z+12x^{3}+24x^{2}y^{2}-120x^{2}yz+72x^{2}y+150x^{2}y^{2}$ $-180x^{2}z+54x^{2}+32xy^{3}-240xy^{2}z+144xy^{2}+600xyz^{2}-720xyz+216xy-500xz^{3}+900xz^{2}$ $-540xz+108x+16y^{4}-160y^{3}z+96y^{3}+600y^{2}z^{2}-720y^{2}z+216y^{2}-1000yz^{3}+1800yz^{2}$ $-1080yz+216y+625z^{4}-1500z^{3}+1350z^{2}-540z+81$

So, $(x+2y-5z+3)^{4}$ expanded form consists of $35$ terms.

So, the answer is $\boxed{35}$

no of terms in (x1 + x2 + x3 + ... +x_m)^n= (n +m -1)!/((n!)(m-1)!),hence no of terms in above question ( n = m = 4),is equal to (4 + 4 -1)!/((4!)(4 - 1)!) i.e. 35

We can think of the no. of terms in an expansion like this: (x + y) ^ 2 has three terms: x^2, y^2 and xy. There are 3 ways to distribute a power of 2 among terms: either one gets both or they get 1 each. The coefficient is irrelevant here.

Using this idea for the given question, we have to find the no. of ways to distribute a power of 4 among 4 terms. The different ways are: 1. Any one term gets all 4 : x^4 for example.. 2. Two terms get 2 each 3. One term gets 2 and two others get 1 each 4. All of them get 1 5. One term gets 3 and another one gets 1

Now, we use combinatorics to find the no. of combinations of each: 1. 4 ways to do this 2. We must pick a pair of terms. Hence 4C2 3. 4C1 * 3C2 4. 1 way to do this 5. 4C1 * 3C1

The sum is 35.

Suppose (a+b+c+d)^{n} If we go down to the basics of binomial expansion, we realize that we have to distribute \boxed{n}
brackets to 4 terms:a,b,c,d OR we have for each term of expansion a^{p}b^{q}c^{r}d^{s} where p+q+r+s=n. Therefore solution to both the problem is \begin{pmatrix} n+ & 4-1 \ 4 & -1 \end{pmatrix}