EF Multinomial Coefficients

Section 23.2 Multinomial Coefficients

Theorem 23.2.1. Trinomial Theorem.

The expansion of the trinomial

(x + y + z)^{n}

is the sum of all possible products

\frac{n!}{i! j! k!} x^{i} y^{j} z^{k},

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where

0 \leq i, j, k \leq n

such that

i + j + k = n .

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Proof idea.

Similarly to the proof of the Binomial Theorem, write

\begin{matrix} (✶) & (x + y + z)^{n} = (x + y + z) (x + y + z) \dots (x + y + z), \end{matrix}

with

n

factors. To expand this out, we generalize the FOIL method: from each factor, choose either

x,

y,

z,

then multiply all your choices together. For any such product, the powers on

x,

y,

and

z

must sum to

n .

To get the final expansion, add the results of all possible such products.

But we can collect terms that have the same exponent on each of

x,

y,

and

z .

How many ways can we form a specific term

x^{i} y^{j} z^{k},

for

0 \leq i, j, k \leq n

such that

i + j + k = n ?

We have

C_{i}^{n}

ways to choose

i

factors from the right-hand side of (✶) from which to take

x,

then

C_{j}^{n - i}

ways to choose

j

factors from which to take

y .

But now from all remaining factors we must choose

z,

and there is only

1

way to do this. So the coefficient on

x^{i} y^{j} z^{k}

(\binom{n}{i}) (\binom{n - i}{j}) = (\frac{n!}{i! (n - i)!}) (\frac{(n - i)!}{j! (n - i - j)!}) = \frac{n!}{i! j! k!} .

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Alternative proof idea.

Use the Binomial Theorem on

(x + (y + z))^{n},

then again on

(y + z)^{k}

for each term

C_{k}^{n} x^{n - k} (y + z)^{k} .

(This would be very tedious!)

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Worked Example 23.2.2. Expanding a trinomial.

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Determine the terms in the expansion of

(2 x + y - 3 z)^{3} .

Solution.

First, rewrite

(2 x + y - 3 z)^{3} = ((2 x) + y + (- 3 z))^{3} .

So the terms in the expansion involve products

(2 x)^{i} y^{j} (- 3 z)^{k} .

We need to account for all triples of exponents

i, j, k

that sum to

3 .

$i$	$j$	$k$	$\frac{n!}{i! j! k!}$	term	simplified
$3$	$0$	$0$	$1$	$(2 x)^{3}$	$8 x^{3}$
$0$	$3$	$0$	$1$	$y^{3}$	$y^{3}$
$0$	$0$	$3$	$1$	$(- 3 x)^{3}$	$- 27 z^{3}$
$2$	$1$	$0$	$3$	$3 (2 x)^{2} y$	$12 x^{2} y$
$2$	$0$	$1$	$3$	$3 (2 x)^{2} (- 3 z)$	$- 36 x^{2} z$
$1$	$2$	$0$	$3$	$3 (2 x) y^{2}$	$6 x y^{2}$
$0$	$2$	$1$	$3$	$3 y^{2} (- 3 z)$	$- 9 y^{2} z$
$1$	$0$	$2$	$3$	$3 (2 x) (- 3 z)^{2}$	$- 54 x z^{2}$
$0$	$1$	$2$	$3$	$3 y (- 3 z)^{2}$	$- 9 y z^{2}$
$1$	$1$	$1$	$3!$	$6 (2 x) y (- 3 z)$	$- 36 x y z$

Collecting this together, we have

\begin{aligned} (2 x + y - 3 z)^{3} \\ = 8 x^{3} + y^{3} - 27 z^{3} + 12 x^{2} y - 36 x^{2} z \\ + 6 x y^{2} - 9 y^{2} z - 54 x z^{2} - 36 x y z . \end{aligned}

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Worked Example 23.2.3. Determining a specific coefficient in a trinomial expansion.

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Determine the coefficient on

x^{5} y^{2} z^{7}

in the expansion of

(x + y + z)^{14} .

Solution.

Here we don’t have any extra contributions to the coefficient from constants inside the trinomial, so using

n = 14,

i = 5,

j = 2,

k = 7,

the coefficient is simply

\frac{14!}{5! 2! 7!} = \frac{14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 2} = 14 \cdot 13 \cdot 11 \cdot 9 \cdot 4 = 72, 072 .

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The pattern of the Binomial Theorem and Trinomial Theorem continues.

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Theorem 23.2.4. Multinomial Theorem.

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The expansion of

(x_{1} + x_{2} + \dots + x_{m})^{n}

is the sum of all possible products

\frac{n!}{i_{1}! i_{2}! \dots i_{m}!} x_{1}^{i_{1}} x_{2}^{i_{2}} \dots x_{m}^{i_{m}},

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where the exponents

i_{1}, i_{2}, \dots, i_{n}

sum to

n .

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Proof idea.

Use the same generalized FOIL method argument as in the Binomial and Trinomial Theorem proofs, and simplify the product of combination formulas obtained.

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Worked Example 23.2.5. Determining a specific coefficient in a multinomial expansion.

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Determine the coefficient on

x^{2} y z^{6}

in the expansion of

(3 x + 2 y + z^{2} + 6)^{8} .

Solution.

Rewriting

(3 x + 2 y + z^{2} + 6)^{8} = ((3 x) + (2 y) + (z^{2}) + 6)^{8},

we see that the four terms in this multinomial are

3 x, 2 y, z^{2}, 6 .

So what we really want to know is the total coefficient on the term involving

(3 x)^{2} (2 y)^{1} (z^{2})^{3} 6^{2} .

The Multinomial Theorem tells us that there will be

\frac{8!}{2! 1! 3! 2!} = 1, 680

such terms in the expansion of the multinomial. Therefore, we obtain the term

(1, 680) (3 x)^{2} (2 y)^{1} (z^{2})^{3} 6^{2} = (1, 088, 640) x^{2} y z^{6}

with a total coefficient of

1, 088, 640 .

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multinomial coefficient: 🔗
a number appearing as a coefficient in the expansion of $(x_{1} + x_{2} + \dots + x_{m})^{n}$
$(\binom{n}{i_{1}, i_{2}, \dots, i_{m}})$: 🔗
the coefficient on the term $x_{1}^{i_{1}} x_{2}^{i_{2}} \dots x_{m}^{i_{m}}$ in the expansion of $(x_{1} + x_{2} + \dots + x_{m})^{n},$ where the exponents $i_{1}, i_{2}, \dots, i_{m}$ must sum to $n$

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Note 23.2.6.

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The Multinomial Theorem tells us $(\binom{n}{i_{1}, i_{2}, \dots, i_{m}}) = \frac{n!}{i_{1}! i_{2}! \dots i_{m}!} .$
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In the case of a binomial expansion $(x_{1} + x_{2})^{n},$ the term $x_{1}^{i_{1}} x_{2}^{i_{2}}$ must have $i_{1} + i_{2} = n,$ or $i_{2} = n - i_{1} .$ The Multinomial Theorem tells us that the coefficient on this term is
$(\binom{n}{i_{1}, i_{2}}) = \frac{n!}{i_{1}! i_{2}!} = \frac{n!}{i_{1}! (n - i_{1})!} = (\binom{n}{i_{1}}) .$
Therefore, in the case $m = 2,$ the Multinomial Theorem reduces to the Binomial Theorem.

Elementary Foundations: An introduction to topics in discrete mathematics

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Section 23.2 Multinomial Coefficients

Theorem 23.2.1. Trinomial Theorem.

Proof idea.

Alternative proof idea.

Worked Example 23.2.2. Expanding a trinomial.

Worked Example 23.2.3. Determining a specific coefficient in a trinomial expansion.

Theorem 23.2.4. Multinomial Theorem.

Proof idea.

Worked Example 23.2.5. Determining a specific coefficient in a multinomial expansion.

Note 23.2.6.