Master Multinomial Theorem - Intermediate-Advanced Level Problems Multinomial Theorem INTERMEDIATE ADVANCED

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y+z)^5
- Desired term: the term x^1 * y^4
- Exponents: x = 1, y = 4, z = 0

Step 1 - Verify exponent sum:
1 + 4 + 0 = 5 = 5 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{5!}{1! × 4!}$

Step 3 - Calculate:
- Numerator: 5! = 120
- Denominator: 1! × 4! = 1 × 24
- Denominator value: 24

Final Calculation:
Coefficient = 120 / 24 = 5

Alternative interpretation: This equals the number of ways to arrange 5 items with:
1 of type x, 4 of type y, 0 of type z

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 3^5 = 243

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y)^7
- Desired term: the term x^5 * y^2
- Exponents: x = 5, y = 2

Step 1 - Verify exponent sum:
5 + 2 = 7 = 7 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{7!}{5! × 2!}$

Step 3 - Calculate:
- Numerator: 7! = 5040
- Denominator: 5! × 2! = 120 × 2
- Denominator value: 240

Final Calculation:
Coefficient = 5040 / 240 = 21

Alternative interpretation: This equals the number of ways to arrange 7 items with:
5 of type x, 2 of type y

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 2^7 = 128

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y)^4
- Desired term: the term x^4
- Exponents: x = 4, y = 0

Step 1 - Verify exponent sum:
4 + 0 = 4 = 4 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{4!}{4!}$

Step 3 - Calculate:
- Numerator: 4! = 24
- Denominator: 4! = 24
- Denominator value: 24

Final Calculation:
Coefficient = 24 / 24 = 1

Alternative interpretation: This equals the number of ways to arrange 4 items with:
4 of type x, 0 of type y

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 2^4 = 16

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y+z)^7
- Desired term: the term y^6 * z^1
- Exponents: x = 0, y = 6, z = 1

Step 1 - Verify exponent sum:
0 + 6 + 1 = 7 = 7 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{7!}{6! × 1!}$

Step 3 - Calculate:
- Numerator: 7! = 5040
- Denominator: 6! × 1! = 720 × 1
- Denominator value: 720

Final Calculation:
Coefficient = 5040 / 720 = 7

Alternative interpretation: This equals the number of ways to arrange 7 items with:
0 of type x, 6 of type y, 1 of type z

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 3^7 = 2187

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y)^4
- Desired term: the term x^3 * y^1
- Exponents: x = 3, y = 1

Step 1 - Verify exponent sum:
3 + 1 = 4 = 4 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{4!}{3! × 1!}$

Step 3 - Calculate:
- Numerator: 4! = 24
- Denominator: 3! × 1! = 6 × 1
- Denominator value: 6

Final Calculation:
Coefficient = 24 / 6 = 4

Alternative interpretation: This equals the number of ways to arrange 4 items with:
3 of type x, 1 of type y

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 2^4 = 16

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y)^6
- Desired term: the term x^1 * y^5
- Exponents: x = 1, y = 5

Step 1 - Verify exponent sum:
1 + 5 = 6 = 6 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{6!}{1! × 5!}$

Step 3 - Calculate:
- Numerator: 6! = 720
- Denominator: 1! × 5! = 1 × 120
- Denominator value: 120

Final Calculation:
Coefficient = 720 / 120 = 6

Alternative interpretation: This equals the number of ways to arrange 6 items with:
1 of type x, 5 of type y

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 2^6 = 64

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y+z)^5
- Desired term: the term x^1 * z^4
- Exponents: x = 1, y = 0, z = 4

Step 1 - Verify exponent sum:
1 + 0 + 4 = 5 = 5 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{5!}{1! × 4!}$

Step 3 - Calculate:
- Numerator: 5! = 120
- Denominator: 1! × 4! = 1 × 24
- Denominator value: 24

Final Calculation:
Coefficient = 120 / 24 = 5

Alternative interpretation: This equals the number of ways to arrange 5 items with:
1 of type x, 0 of type y, 4 of type z

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 3^5 = 243

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y+z)^4
- Desired term: the term x^2 * y^2
- Exponents: x = 2, y = 2, z = 0

Step 1 - Verify exponent sum:
2 + 2 + 0 = 4 = 4 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{4!}{2! × 2!}$

Step 3 - Calculate:
- Numerator: 4! = 24
- Denominator: 2! × 2! = 2 × 2
- Denominator value: 4

Final Calculation:
Coefficient = 24 / 4 = 6

Alternative interpretation: This equals the number of ways to arrange 4 items with:
2 of type x, 2 of type y, 0 of type z

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 3^4 = 81

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y+z)^4
- Desired term: the term x^4
- Exponents: x = 4, y = 0, z = 0

Step 1 - Verify exponent sum:
4 + 0 + 0 = 4 = 4 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{4!}{4!}$

Step 3 - Calculate:
- Numerator: 4! = 24
- Denominator: 4! = 24
- Denominator value: 24

Final Calculation:
Coefficient = 24 / 24 = 1

Alternative interpretation: This equals the number of ways to arrange 4 items with:
4 of type x, 0 of type y, 0 of type z

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 3^4 = 81

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y)^4
- Desired term: the term y^4
- Exponents: x = 0, y = 4

Step 1 - Verify exponent sum:
0 + 4 = 4 = 4 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{4!}{4!}$

Step 3 - Calculate:
- Numerator: 4! = 24
- Denominator: 4! = 24
- Denominator value: 24

Final Calculation:
Coefficient = 24 / 24 = 1

Alternative interpretation: This equals the number of ways to arrange 4 items with:
0 of type x, 4 of type y

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 2^4 = 16

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y+z)^6
- Desired term: the term x^1 * y^1 * z^4
- Exponents: x = 1, y = 1, z = 4

Step 1 - Verify exponent sum:
1 + 1 + 4 = 6 = 6 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{6!}{1! × 1! × 4!}$

Step 3 - Calculate:
- Numerator: 6! = 720
- Denominator: 1! × 1! × 4! = 1 × 1 × 24
- Denominator value: 24

Final Calculation:
Coefficient = 720 / 24 = 30

Alternative interpretation: This equals the number of ways to arrange 6 items with:
1 of type x, 1 of type y, 4 of type z

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 3^6 = 729

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y)^6
- Desired term: the term x^6
- Exponents: x = 6, y = 0

Step 1 - Verify exponent sum:
6 + 0 = 6 = 6 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{6!}{6!}$

Step 3 - Calculate:
- Numerator: 6! = 720
- Denominator: 6! = 720
- Denominator value: 720

Final Calculation:
Coefficient = 720 / 720 = 1

Alternative interpretation: This equals the number of ways to arrange 6 items with:
6 of type x, 0 of type y

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 2^6 = 64

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y+z)^5
- Desired term: the term z^5
- Exponents: x = 0, y = 0, z = 5

Step 1 - Verify exponent sum:
0 + 0 + 5 = 5 = 5 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{5!}{5!}$

Step 3 - Calculate:
- Numerator: 5! = 120
- Denominator: 5! = 120
- Denominator value: 120

Final Calculation:
Coefficient = 120 / 120 = 1

Alternative interpretation: This equals the number of ways to arrange 5 items with:
0 of type x, 0 of type y, 5 of type z

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 3^5 = 243

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y)^5
- Desired term: the term x^5
- Exponents: x = 5, y = 0

Step 1 - Verify exponent sum:
5 + 0 = 5 = 5 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{5!}{5!}$

Step 3 - Calculate:
- Numerator: 5! = 120
- Denominator: 5! = 120
- Denominator value: 120

Final Calculation:
Coefficient = 120 / 120 = 1

Alternative interpretation: This equals the number of ways to arrange 5 items with:
5 of type x, 0 of type y

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 2^5 = 32

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y+z)^6
- Desired term: the term x^4 * y^1 * z^1
- Exponents: x = 4, y = 1, z = 1

Step 1 - Verify exponent sum:
4 + 1 + 1 = 6 = 6 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{6!}{4! × 1! × 1!}$

Step 3 - Calculate:
- Numerator: 6! = 720
- Denominator: 4! × 1! × 1! = 24 × 1 × 1
- Denominator value: 24

Final Calculation:
Coefficient = 720 / 24 = 30

Alternative interpretation: This equals the number of ways to arrange 6 items with:
4 of type x, 1 of type y, 1 of type z

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 3^6 = 729

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y+z)^5
- Desired term: the term x^3 * y^2
- Exponents: x = 3, y = 2, z = 0

Step 1 - Verify exponent sum:
3 + 2 + 0 = 5 = 5 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{5!}{3! × 2!}$

Step 3 - Calculate:
- Numerator: 5! = 120
- Denominator: 3! × 2! = 6 × 2
- Denominator value: 12

Final Calculation:
Coefficient = 120 / 12 = 10

Alternative interpretation: This equals the number of ways to arrange 5 items with:
3 of type x, 2 of type y, 0 of type z

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 3^5 = 243

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y+z)^5
- Desired term: the term x^1 * z^4
- Exponents: x = 1, y = 0, z = 4

Step 1 - Verify exponent sum:
1 + 0 + 4 = 5 = 5 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{5!}{1! × 4!}$

Step 3 - Calculate:
- Numerator: 5! = 120
- Denominator: 1! × 4! = 1 × 24
- Denominator value: 24

Final Calculation:
Coefficient = 120 / 24 = 5

Alternative interpretation: This equals the number of ways to arrange 5 items with:
1 of type x, 0 of type y, 4 of type z

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 3^5 = 243

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y+z)^4
- Desired term: the term y^3 * z^1
- Exponents: x = 0, y = 3, z = 1

Step 1 - Verify exponent sum:
0 + 3 + 1 = 4 = 4 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{4!}{3! × 1!}$

Step 3 - Calculate:
- Numerator: 4! = 24
- Denominator: 3! × 1! = 6 × 1
- Denominator value: 6

Final Calculation:
Coefficient = 24 / 6 = 4

Alternative interpretation: This equals the number of ways to arrange 4 items with:
0 of type x, 3 of type y, 1 of type z

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 3^4 = 81

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y)^6
- Desired term: the term x^3 * y^3
- Exponents: x = 3, y = 3

Step 1 - Verify exponent sum:
3 + 3 = 6 = 6 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{6!}{3! × 3!}$

Step 3 - Calculate:
- Numerator: 6! = 720
- Denominator: 3! × 3! = 6 × 6
- Denominator value: 36

Final Calculation:
Coefficient = 720 / 36 = 20

Alternative interpretation: This equals the number of ways to arrange 6 items with:
3 of type x, 3 of type y

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 2^6 = 64

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y+z)^5
- Desired term: the term x^2 * y^3
- Exponents: x = 2, y = 3, z = 0

Step 1 - Verify exponent sum:
2 + 3 + 0 = 5 = 5 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{5!}{2! × 3!}$

Step 3 - Calculate:
- Numerator: 5! = 120
- Denominator: 2! × 3! = 2 × 6
- Denominator value: 12

Final Calculation:
Coefficient = 120 / 12 = 10

Alternative interpretation: This equals the number of ways to arrange 5 items with:
2 of type x, 3 of type y, 0 of type z

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 3^5 = 243