Problem 31 - Algebra 2 and Trigonometry

Question

A box contains 9 red, 4 blue, and 6 vellow chips. In how many ways can 6 chips be chosen if: a. all 6 chips are red? c. 2 chips are blue? e. 4 chips are yellow? g. 3 chips are red and 3 chips are blue? b. all 6 chips are yellow? d. 3 chips are red? f. there are 2 chips of each color? h. 5 chips are red and 1 chip is yellow?

Step-by-Step Solution

Answer

a. 84, c. 6, e. 15, g. 336, b. 1, d. 84, f. 3240, h. 756.

Step 1: Calculate Combination Formula

We use the combination formula \( \binom{n}{r} \) to determine the number of ways to choose \( r \) items from \( n \) items without regard to the order of selection. The formula is given by \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \).

Step 2: Solve Part a

For part (a), we want to choose all 6 chips from the red ones. There are 9 red chips.Use the combination formula: \( \binom{9}{6} = \frac{9!}{6!3!} = 84 \) So, there are 84 ways to choose 6 red chips.

Step 3: Solve Part b

For part (b), attempt to choose 6 chips from the yellow ones. There are only 6 yellow chips.Use the combination formula: \( \binom{6}{6} = 1 \)Thus, there is exactly 1 way to choose all yellow chips.

Step 4: Solve Part c

For part (c), we need to choose 2 blue chips from the 4 available.Use the combination formula: \( \binom{4}{2} = \frac{4!}{2!2!} = 6 \)Therefore, there are 6 ways to choose 2 blue chips.

Step 5: Solve Part d

For part (d), choose 3 red chips from 9.Use the combination formula: \( \binom{9}{3} = \frac{9!}{3!6!} = 84 \)Thus, there are 84 ways to choose 3 red chips.

Step 6: Solve Part e

For part (e), choose 4 yellow chips from 6 available.Use the combination formula: \( \binom{6}{4} = \frac{6!}{4!2!} = 15 \)Hence, there are 15 ways to choose 4 yellow chips.

Step 7: Solve Part f

For part (f), choose 2 chips of each color.Calculate combinations for each color: - Red: \( \binom{9}{2} = 36 \) - Blue: \( \binom{4}{2} = 6 \) - Yellow: \( \binom{6}{2} = 15 \) Multiply the results: \( 36 \times 6 \times 15 = 3240 \)Therefore, there are 3240 ways to choose 2 chips of each color.

Step 8: Solve Part g

For part (g), choose 3 red and 3 blue chips.Calculate combinations for each color:- Red: \( \binom{9}{3} = 84 \) - Blue: \( \binom{4}{3} = 4 \) Multiply the results: \( 84 \times 4 = 336 \)Thus, there are 336 ways to choose 3 red chips and 3 blue chips.

Step 9: Solve Part h

For part (h), choose 5 red chips and 1 yellow chip.Calculate combinations for each color:- Red: \( \binom{9}{5} = 126 \) - Yellow: \( \binom{6}{1} = 6 \) Multiply the results: \( 126 \times 6 = 756 \)Thus, there are 756 ways to choose 5 red chips and 1 yellow chip.

Key Concepts

Combination Formula

In combinatorics, the combination formula is key to finding out how many different ways we can choose items from a larger set, without caring about the order. Imagine you have a unique set of items and you want to select a few of them. But, unlike permutations, the sequence or arrangement of these selected items doesn't matter. This is where combinations come into play. The mathematical formula for combinations, represented as \( \binom{n}{r} \), helps to calculate this. It is expressed as:\[\binom{n}{r} = \frac{n!}{r!(n-r)!}\]Here, \( n \) stands for the total number of items available, and \( r \) is the number of items you wish to select. The exclamation mark \(!\) indicates factorial, which means multiplying a series of descending natural numbers. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
Factors of \( n \) above \( r \) are divided by corresponding factors below \( r \). This simplicity makes calculating combinations very efficient, even for large sets of items.

Probability

Probability is all about determining how likely an event is to happen. When applying in scenarios involving chips, such as choosing certain colored chips from a box, it can provide significant insights. How we can link probability to combinations is through calculating the outcome possibilities where certain conditions are met.
Let's take two chips as an example: Suppose there are different colored chips in a box, and you want to determine the probability of selecting 2 blue ones. If there are 4 blue chips in total, using the combination formula, there are 6 possible ways to select any 2 blue chips out of 4. To find the probability, you'd compare this to the total number of ways to select any 2 chips from the entire batch.
Thus, probability hinges on the relation of successful outcomes to total possible outcomes, making the understanding of probability much more intuitive with the use of combination.

Binomial Coefficients

Binomial coefficients form the backbone of combinations, and they appear in the mystical world of binomial expansions. If you've ever come across the binomial theorem, you'd recognize them as the numbers in the expansion of \((a+b)^n\). These coefficients don't just indicate how elements are to be combined in an equation; they give insight into every possible outcome when performing selections.
The term "binomial coefficients" refers to the numbers obtained from Pascal's triangle, where each number is the sum of the two directly above it. The combination formula \( \binom{n}{r} \) is essentially a binomial coefficient representing how many ways \( r \) items can be chosen from \( n \).
In practical scenarios, these coefficients help us structure selections in tasks like choosing colored chips from a box without worrying about order but focusing on quantity and condition fulfillment.
For example, when selecting 3 red chips to ensure a specific outcome, the binomial coefficient indicates the count of potential combinations, paving the way to complete calculation strings in combinatorial ways.