Why is the expansion of $(1-x)^{-4}$ using $\binom{n+3}{3}$?

This is the generalized binomial theorem for negative exponents. The coefficient of $x^n$ in $(1-x)^{-r}$ is given by $\binom{n+r-1}{r-1}$. Here, $r=4$, so the formula becomes $\binom{n+4-1}{4-1} = \binom{n+3}{3}$. It is a powerful shortcut that saves you from tedious manual expansion.

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Animated Solution for Mathematics - Permutations and Combinations: The number of ways of getting a sum 16 on throwing a dice four times is______

Q: Where did the $x^4$ go in the final calculation?

Great observation! Since we factored out $x^4$ from the expression, we shifted our target. Instead of finding the coefficient of $x^{16}$ in the original expression, we now look for the coefficient of $x^{16-4} = x^{12}$ in the remaining part. It is a mathematical 'shift' to make the expansion easier to handle.

Q: Why did we stop the expansion of $(1-x^6)^4$ at $6x^{12}$?

Because we only need the coefficient of $x^{12}$. Any term higher than $x^{12}$ in the expansion of $(1-x^6)^4$ would require a power of $x$ from the $(1-x)^{-4}$ part that is negative, which doesn't exist in our series. We only care about terms that contribute to $x^{12}$.

Enter Numerical Value:

Visualized Solution

Defining the Problem Equation

Let the outcomes of the four throws be $x_{1}, x_{2}, x_{3}, x_{4}$ .
We need to find the number of solutions to: $x_{1} + x_{2} + x_{3} + x_{4} = 16$ .
Constraint for each throw: $1 \leq x_{i} \leq 6$ for $i = 1, 2, 3, 4$ .

The Generating Function Approach

The generating function for one die is $(x^{1} + x^{2} + x^{3} + x^{4} + x^{5} + x^{6})$ .
This polynomial captures all possible outcomes of a single throw.

Generating Function for Four Dice

For four dice, the function is $(x^{1} + x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{4}$ .
The number of ways is the coefficient of $x^{16}$ in this expansion.

Simplifying the Expression

Factor out $x$ from the bracket: $(x (1 + x + x^{2} + x^{3} + x^{4} + x^{5}))^{4}$ .
Sum of GP: $1 + x + x^{2} + x^{3} + x^{4} + x^{5} = \frac{1 - x ^{6}}{1 - x}$ .
The expression becomes: $x^{4} (\frac{1 - x ^{6}}{1 - x})^{4}$ .

Shifting the Target Power

Expression: $x^{4} (1 - x^{6})^{4} (1 - x)^{- 4}$ .
Since we have $x^{4}$ outside, we need the coefficient of $x^{12}$ in $(1 - x^{6})^{4} (1 - x)^{- 4}$ .

Expanding the Numerator

Expansion: $(1 - x^{6})^{4} = (0 4) - (1 4) x^{6} + (2 4) x^{12} - \dots$
Simplified: $1 - 4 x^{6} + 6 x^{12} - \dots$

Negative Binomial Expansion Tool

Formula: Coeff. of $x^{n}$ in $(1 - x)^{- r}$ is $(r - 1 n + r - 1)$ .
For $r = 4$ , Coeff. of $x^{n}$ is $(4 - 1 n + 4 - 1) = (3 n + 3)$ .

Identifying Contributing Terms

Total Coeff. of $x^{12} = 1 \cdot (Coeff. of x^{12}) - 4 \cdot (Coeff. of x^{6}) + 6 \cdot (Coeff. of x^{0})$
Using $(3 n + 3)$ : $1 \cdot (3 12 + 3) - 4 \cdot (3 6 + 3) + 6 \cdot (3 0 + 3)$

Calculating the First Term

First term: $(3 15) = \frac{15 \times 14 \times 13}{3 \times 2 \times 1}$
$(3 15) = 5 \times 7 \times 13 = 455$

Calculating the Second Term

Second term: $4 \times (3 9) = 4 \times \frac{9 \times 8 \times 7}{3 \times 2 \times 1}$
$4 \times (3 9) = 4 \times (3 \times 4 \times 7) = 4 \times 84 = 336$

Calculating the Third Term

Third term: $6 \times (3 3) = 6 \times 1 = 6$

Final Summation

Total ways $= 455 - 336 + 6$
Total ways $= 119 + 6 = 125$

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The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Setup

To determine the number of ways to obtain a sum of 16 with four standard six-sided dice, we define the outcome of each die as

x_{i}

, where

1 \leq x_{i} \leq 6

. We seek the number of integer solutions to the equation:

x_{1} + x_{2} + x_{3} + x_{4} = 16

Since each die is constrained to a finite range, we utilize the power of generating functions. The possible outcomes for a single die are represented by the polynomial:

P (x) = x^{1} + x^{2} + x^{3} + x^{4} + x^{5} + x^{6}

The total number of ways to achieve a sum

S

is the coefficient of

x^{S}

in the expansion of

[P (x)]^{4}

. Therefore, we must find the coefficient of

x^{16}

in the expression

(x^{1} + x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{4}

The Master Equation

We simplify the expression by factoring out

x

from the polynomial:

x^{4} (1 + x + x^{2} + x^{3} + x^{4} + x^{5})^{4}

Recognizing the term inside the parentheses as a finite geometric series, we apply the identity

1 + x + x^{2} + x^{3} + x^{4} + x^{5} = \frac{1 - x ^{6}}{1 - x}

. Substituting this into our expression yields:

x^{4} (\frac{1 - x ^{6}}{1 - x})^{4} = x^{4} (1 - x^{6})^{4} (1 - x)^{- 4}

Because of the

x^{4}

factor, finding the coefficient of

x^{16}

in the original expression is equivalent to finding the coefficient of

x^{12}

in the expansion of

(1 - x^{6})^{4} (1 - x)^{- 4}

The Expansion Process

We expand the two components of our product. First, using the binomial theorem for the numerator:

(1 - x^{6})^{4} = (0 4) - (1 4) x^{6} + (2 4) x^{12} - \dots = 1 - 4 x^{6} + 6 x^{12} - \dots

Next, we use the negative binomial expansion for

(1 - x)^{- 4}

. The general term for

(1 - x)^{- r}

(r - 1 n + r - 1) x^{n}

. With

r = 4

, the coefficient of

x^{n}

(3 n + 3)

Final Calculation

To find the coefficient of

x^{12}

in the product

(1 - 4 x^{6} + 6 x^{12}) (1 - x)^{- 4}

, we distribute the terms:

1 \times (coefficient of x^{12} in (1 - x)^{- 4}) = 1 \times (3 12 + 3) = (3 15) = 455

- 4 x^{6} \times (coefficient of x^{6} in (1 - x)^{- 4}) = - 4 \times (3 6 + 3) = - 4 \times 84 = - 336

6 x^{12} \times (coefficient of x^{0} in (1 - x)^{- 4}) = 6 \times (3 0 + 3) = 6 \times 1 = 6

Summing these contributions together, we arrive at the final result:

455 - 336 + 6 = 125

There are exactly 125 ways to roll a sum of 16 with four six-sided dice.

Common Traps & FAQs

Why can't we just use the stars and bars formula

(k - 1 n + k - 1)

directly?

The stars and bars formula assumes there is no upper limit on the number of items in each bin. Here, each die is capped at 6. If you use stars and bars blindly, you include cases like (10, 2, 2, 2), which is impossible for a standard die. We use generating functions to 'filter out' these invalid cases.

Where did the

x^{4}

go in the final calculation?

Great observation! Since we factored out

x^{4}

from the expression, we shifted our target. Instead of finding the coefficient of

x^{16}

in the original expression, we now look for the coefficient of

x^{16 - 4} = x^{12}

in the remaining part. It is a mathematical 'shift' to make the expansion easier to handle.

Why is the expansion of

(1 - x)^{- 4}

using

(3 n + 3)

This is the generalized binomial theorem for negative exponents. The coefficient of

x^{n}

(1 - x)^{- r}

is given by

(r - 1 n + r - 1)

. Here,

r = 4

, so the formula becomes

(4 - 1 n + 4 - 1) = (3 n + 3)

. It is a powerful shortcut that saves you from tedious manual expansion.

Is there any other way to solve this without generating functions?

Yes, you could use the Principle of Inclusion-Exclusion (PIE). You would take the total ways (stars and bars) and subtract cases where at least one die exceeds 6, add back cases where two dice exceed 6, and so on. Generating functions are just a more elegant, systematic way of performing PIE.

Why did we stop the expansion of

(1 - x^{6})^{4}

6 x^{12}

Because we only need the coefficient of

x^{12}

. Any term higher than

x^{12}

in the expansion of

(1 - x^{6})^{4}

would require a power of

x

from the

(1 - x)^{- 4}

part that is negative, which doesn't exist in our series. We only care about terms that contribute to

x^{12}

Animated Solution for Mathematics - Permutations and Combinations: The number of ways of getting a sum 16 on throwing a dice four times is______

Enter Numerical Value:

Visualized Solution

Defining the Problem Equation

The Generating Function Approach

Generating Function for Four Dice

Simplifying the Expression

Shifting the Target Power

Expanding the Numerator

Negative Binomial Expansion Tool

Identifying Contributing Terms

Calculating the First Term

Calculating the Second Term

Calculating the Third Term

Final Summation

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Master Equation

The Expansion Process

Final Calculation

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