Simplify And Evaluate The Expression:(1) ( 4 ) 4 × ( 4 2 ) × ( 2 ) 3 × ( 2 ) 5 (4)^4 \times (4^2) \times (2)^3 \times (2)^5 ( 4 ) 4 × ( 4 2 ) × ( 2 ) 3 × ( 2 ) 5 ( 4 5 3 2 ) 3 = ? \binom{4^5}{3^2}^3 = ? ( 3 2 4 5 ​ ) 3 = ?

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Introduction

In this article, we will simplify and evaluate the given mathematical expression. The expression involves exponentiation, multiplication, and combination. We will break down the expression step by step and simplify it using the rules of exponents and combination.

Understanding the Expression

The given expression is:

(4)4×(42)×(2)3×(2)5(4532)3(4)^4 \times (4^2) \times (2)^3 \times (2)^5 \binom{4^5}{3^2}^3

This expression involves several operations:

  • Exponentiation: (4)4(4)^4, (42)(4^2), (2)3(2)^3, and (2)5(2)^5
  • Multiplication: (4)4×(42)(4)^4 \times (4^2) and (2)3×(2)5(2)^3 \times (2)^5
  • Combination: (4532)3\binom{4^5}{3^2}^3

Simplifying Exponents

To simplify the expression, we will start by simplifying the exponents.

  • (4)4(4)^4 can be simplified as 44=2564^4 = 256
  • (42)(4^2) can be simplified as 42=164^2 = 16
  • (2)3(2)^3 can be simplified as 23=82^3 = 8
  • (2)5(2)^5 can be simplified as 25=322^5 = 32

Simplifying Multiplication

Now that we have simplified the exponents, we can simplify the multiplication.

  • (4)4×(42)(4)^4 \times (4^2) can be simplified as 256×16=4096256 \times 16 = 4096
  • (2)3×(2)5(2)^3 \times (2)^5 can be simplified as 8×32=2568 \times 32 = 256

Understanding Combination

The combination (4532)3\binom{4^5}{3^2}^3 can be simplified using the formula for combination:

(nr)=n!r!(n−r)!\binom{n}{r} = \frac{n!}{r!(n-r)!}

In this case, n=45n = 4^5 and r=32r = 3^2.

Simplifying Combination

To simplify the combination, we need to calculate 454^5 and 323^2.

  • 45=10244^5 = 1024
  • 32=93^2 = 9

Now, we can substitute these values into the combination formula:

(10249)=1024!9!(1024−9)!\binom{1024}{9} = \frac{1024!}{9!(1024-9)!}

However, this is not a straightforward calculation. We can simplify the combination using the property of combination:

(nr)=(n−1r−1)+(n−1r)\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}

Using this property, we can simplify the combination as follows:

(10249)=(10238)+(10239)\binom{1024}{9} = \binom{1023}{8} + \binom{1023}{9}

We can continue simplifying the combination using this property until we reach a manageable value.

Simplifying Combination Using Property

Using the property of combination, we can simplify the combination as follows:

(10249)=(10238)+(10239)\binom{1024}{9} = \binom{1023}{8} + \binom{1023}{9}

=(10227)+(10228)+(10239)= \binom{1022}{7} + \binom{1022}{8} + \binom{1023}{9}

=(10216)+(10217)+(10228)+(10239)= \binom{1021}{6} + \binom{1021}{7} + \binom{1022}{8} + \binom{1023}{9}

=(10205)+(10206)+(10217)+(10228)+(10239)= \binom{1020}{5} + \binom{1020}{6} + \binom{1021}{7} + \binom{1022}{8} + \binom{1023}{9}

=(10194)+(10195)+(10206)+(10217)+(10228)+(10239)= \binom{1019}{4} + \binom{1019}{5} + \binom{1020}{6} + \binom{1021}{7} + \binom{1022}{8} + \binom{1023}{9}

=(10183)+(10184)+(10195)+(10206)+(10217)+(10228)+(10239)= \binom{1018}{3} + \binom{1018}{4} + \binom{1019}{5} + \binom{1020}{6} + \binom{1021}{7} + \binom{1022}{8} + \binom{1023}{9}

=(10172)+(10173)+(10184)+(10195)+(10206)+(10217)+(10228)+(10239)= \binom{1017}{2} + \binom{1017}{3} + \binom{1018}{4} + \binom{1019}{5} + \binom{1020}{6} + \binom{1021}{7} + \binom{1022}{8} + \binom{1023}{9}

=(10161)+(10162)+(10173)+(10184)+(10195)+(10206)+(10217)+(10228)+(10239)= \binom{1016}{1} + \binom{1016}{2} + \binom{1017}{3} + \binom{1018}{4} + \binom{1019}{5} + \binom{1020}{6} + \binom{1021}{7} + \binom{1022}{8} + \binom{1023}{9}

=(10150)+(10151)+(10162)+(10173)+(10184)+(10195)+(10206)+(10217)+(10228)+(10239)= \binom{1015}{0} + \binom{1015}{1} + \binom{1016}{2} + \binom{1017}{3} + \binom{1018}{4} + \binom{1019}{5} + \binom{1020}{6} + \binom{1021}{7} + \binom{1022}{8} + \binom{1023}{9}

=1+(10151)+(10162)+(10173)+(10184)+(10195)+(10206)+(10217)+(10228)+(10239)= 1 + \binom{1015}{1} + \binom{1016}{2} + \binom{1017}{3} + \binom{1018}{4} + \binom{1019}{5} + \binom{1020}{6} + \binom{1021}{7} + \binom{1022}{8} + \binom{1023}{9}

Simplifying Combination Using Formula

Now that we have simplified the combination using the property of combination, we can simplify it using the formula for combination:

(nr)=n!r!(n−r)!\binom{n}{r} = \frac{n!}{r!(n-r)!}

In this case, n=1023n = 1023 and r=9r = 9.

(10239)=1023!9!(1023−9)!\binom{1023}{9} = \frac{1023!}{9!(1023-9)!}

=1023!9!×1014!= \frac{1023!}{9! \times 1014!}

=1023×1022×1021×1020×1019×1018×1017×1016×10159×8×7×6×5×4×3×2×1= \frac{1023 \times 1022 \times 1021 \times 1020 \times 1019 \times 1018 \times 1017 \times 1016 \times 1015}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}

=1023×1022×1021×1020×1019×1018×1017×1016×1015362880= \frac{1023 \times 1022 \times 1021 \times 1020 \times 1019 \times 1018 \times 1017 \times 1016 \times 1015}{362880}

=121645100408832000= 121645100408832000

Simplifying Expression

Now that we have simplified the combination, we can simplify the expression:

(4)4×(42)×(2)3×(2)5(4532)3(4)^4 \times (4^2) \times (2)^3 \times (2)^5 \binom{4^5}{3^2}^3

=4096×16×256×121645100408832000= 4096 \times 16 \times 256 \times 121645100408832000

=121645100408832000×4096×16×256= 121645100408832000 \times 4096 \times 16 \times 256

=121645100408832000×2097152= 121645100408832000 \times 2097152

=2551551111111111111112= 2551551111111111111112

The final answer is 2551551111111111111112\boxed{2551551111111111111112}.

Introduction

In our previous article, we simplified and evaluated the mathematical expression (4)4×(42)×(2)3×(2)5(4532)3(4)^4 \times (4^2) \times (2)^3 \times (2)^5 \binom{4^5}{3^2}^3. In this article, we will answer some frequently asked questions related to simplifying and evaluating mathematical expressions.

Q: What is the order of operations in mathematics?

A: The order of operations in mathematics is a set of rules that tells us which operations to perform first when we have multiple operations in an expression. The order of operations is:

  1. Parentheses: Evaluate expressions inside parentheses first.
  2. Exponents: Evaluate any exponential expressions next.
  3. Multiplication and Division: Evaluate any multiplication and division operations from left to right.
  4. Addition and Subtraction: Finally, evaluate any addition and subtraction operations from left to right.

Q: How do I simplify an expression with exponents?

A: To simplify an expression with exponents, we need to follow the rules of exponents. The rules of exponents are:

  • am×an=am+na^m \times a^n = a^{m+n}
  • (am)n=am×n(a^m)^n = a^{m \times n}
  • am÷an=am−na^m \div a^n = a^{m-n}

Q: How do I simplify a combination?

A: To simplify a combination, we need to use the formula for combination:

(nr)=n!r!(n−r)!\binom{n}{r} = \frac{n!}{r!(n-r)!}

We can also use the property of combination to simplify the combination:

(nr)=(n−1r−1)+(n−1r)\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}

Q: What is the difference between a combination and a permutation?

A: A combination is a selection of items from a larger set, where the order of the items does not matter. A permutation is a selection of items from a larger set, where the order of the items does matter.

Q: How do I evaluate an expression with multiple operations?

A: To evaluate an expression with multiple operations, we need to follow the order of operations. We need to evaluate any expressions inside parentheses first, then any exponential expressions, then any multiplication and division operations, and finally any addition and subtraction operations.

Q: What are some common mistakes to avoid when simplifying and evaluating mathematical expressions?

A: Some common mistakes to avoid when simplifying and evaluating mathematical expressions include:

  • Not following the order of operations
  • Not simplifying expressions with exponents
  • Not using the formula for combination
  • Not using the property of combination
  • Not evaluating expressions with multiple operations correctly

Q: How can I practice simplifying and evaluating mathematical expressions?

A: You can practice simplifying and evaluating mathematical expressions by working through practice problems and exercises. You can also try simplifying and evaluating expressions on your own, and then checking your work with a calculator or a teacher.

Q: What are some real-world applications of simplifying and evaluating mathematical expressions?

A: Simplifying and evaluating mathematical expressions has many real-world applications, including:

  • Science: Simplifying and evaluating mathematical expressions is used in science to model and analyze complex systems.
  • Engineering: Simplifying and evaluating mathematical expressions is used in engineering to design and optimize systems.
  • Finance: Simplifying and evaluating mathematical expressions is used in finance to model and analyze financial systems.
  • Computer Science: Simplifying and evaluating mathematical expressions is used in computer science to develop algorithms and data structures.

The final answer is that simplifying and evaluating mathematical expressions is an important skill that has many real-world applications. By following the order of operations, simplifying expressions with exponents, and using the formula for combination, we can simplify and evaluate mathematical expressions with ease.