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Chapter 10: Problem 64

Use a calculator to verify that each pair of combinations is equal. $$_{7} C_{2},_{7} C_{5}$$

Short Answer

Expert verified
Both \(_{7}C_{2}\) and \(_{7}C_{5}\) are equal to 21.

Step by step solution

01

Understand the Combination Formula

The formula for combinations is given by \[ _nC_r = \frac{n!}{r!(n-r)!} \]where \(!\) denotes factorial, \(n\) is the total number of items, and \(r\) is the number of items to choose.
02

Calculate \(_{7}C_{2}\) Using the Formula

Substituting into the formula, we have:\[_{7}C_{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2! \times 5!}\]Calculating further, \(7! = 7 \times 6 \times 5!\), so:\[_{7}C_{2} = \frac{7 \times 6 \times 5!}{2! \times 5!} = \frac{7 \times 6}{2!}\]\(2! = 2 \times 1 = 2\), so:\[_{7}C_{2} = \frac{42}{2} = 21\]
03

Calculate \(_{7}C_{5}\) Using the Formula

Using the same method for \(_{7}C_{5}\):\[_{7}C_{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5! \times 2!}\]Calculating further:\[_{7}C_{5} = \frac{7 \times 6 \times 5!}{5! \times 2!} = \frac{7 \times 6}{2!}\]As in the previous step, \(2! = 2 \times 1 = 2\), so:\[_{7}C_{5} = \frac{42}{2} = 21\]
04

Verify that the Combinations are Equal

Both \(_{7}C_{2}\) and \(_{7}C_{5}\) calculated to be 21. Since both expressions result in the same value, the combinations are indeed equal, demonstrating the symmetry property of combinations where \(_nC_r = _nC_{n-r}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorial
The factorial of a number is a key concept in combinations. It is denoted by an exclamation mark (!). A factorial represents the product of an integer and all the integers below it. For example, the factorial of 5, written as \(5!\), is calculated as:

\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]

This pattern continues with larger numbers. The formula is useful in permutations and combinations as it helps count the number of possible arrangements in a set.

  • \(0!\) is defined as 1.
  • Factorials grow rapidly with larger numbers.

Understanding factorials is crucial for solving combination and permutation problems efficiently.
Symmetric Property
The symmetric property in combinations is an interesting and useful aspect. It states that choosing \(r\) items from \(n\) is the same as choosing \(n-r\) items from \(n\). This is symbolically represented as:

\[_nC_r = _nC_{n-r}\]

Using this property can simplify calculations, especially with large numbers. In our exercise, we saw this with \(_{7}C_{2} = _{7}C_{5}\), because choosing 2 out of 7 is the same as leaving 5 behind:

  • This reduces the work when calculating combinations for larger sets.
  • It's particularly helpful in verifying equal combinations quickly.

The symmetric property is a powerful tool in combinatorial mathematics.
Combination Formula
The combination formula is essential for determining the number of ways to choose \(r\) items from a set of \(n\) items without regard to order. This is calculated using:

\[_nC_r = \frac{n!}{r!(n-r)!}\]

This formula accounts for different arrangements by dividing by the factorial of \(r\) and \(n-r\). It's only concerned with the selection, not the sequence, distinguishing it from permutations.

  • Useful in probability and statistics.
  • Helps in planning and decision-making where selection matters.

Understanding this formula helps solve problems involving groups and selections easily.

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