Understanding the Problem: Placing Indistinguishable Balls in Distinguishable Bins

The question asks us to find the number of ways to place 8 indistinguishable balls into four distinguishable bins. This is a classic problem in combinatorics, specifically dealing with combinations with repetition. Since the balls are indistinguishable (identical) and the bins are distinguishable (different), the order of placing balls within a bin doesn't matter, and the bins themselves are distinct.

Applying the Stars and Bars Method

Problems involving distributing indistinguishable items into distinguishable containers can be solved using the stars and bars method. Imagine the 8 indistinguishable balls as 'stars' (\(\ast\)). We need to divide these 8 stars into 4 bins. To do this, we can use 'bars' (\(|\)) to separate the bins. If we have 4 bins, we need 3 bars to create partitions between them.

For example, if we have 8 stars and 3 bars, an arrangement like \(\ast\ast | \ast\ast\ast | \ast | \ast\ast\) would mean:

• Bin 1 gets 2 balls
• Bin 2 gets 3 balls
• Bin 3 gets 1 ball
• Bin 4 gets 2 balls

The total number of items to arrange is the number of stars plus the number of bars. In this case, it's 8 stars and \(4-1=3\) bars, totaling \(8+3 = 11\) items.

The number of ways to place the 8 indistinguishable balls into 4 distinguishable bins is equivalent to the number of ways to arrange these 11 items (8 stars and 3 bars). This is a combination problem, as the arrangement is determined by choosing the positions for the bars (or the stars) within the total sequence of 11 positions.

Calculating the Number of Ways

The formula for placing \(n\) indistinguishable items into \(k\) distinguishable bins is given by \(\binom{n+k-1}{k-1}\) or equivalently \(\binom{n+k-1}{n}\).

Here, we have \(n = 8\) (indistinguishable balls) and \(k = 4\) (distinguishable bins).

Using the formula, the number of ways is:

\[ \binom{n+k-1}{k-1} = \binom{8+4-1}{4-1} = \binom{11}{3} \]Or

\[ \binom{n+k-1}{n} = \binom{8+4-1}{8} = \binom{11}{8} \]

Let's calculate \(\binom{11}{3}\):

\[ \binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11!}{3!8!} = \frac{11 \times 10 \times 9 \times 8!}{ (3 \times 2 \times 1) \times 8! } \]

We can cancel out 8! from the numerator and denominator:

\[ \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165 \]

Alternatively, calculating \(\binom{11}{8}\):

\[ \binom{11}{8} = \frac{11!}{8!(11-8)!} = \frac{11!}{8!3!} \]

This is the same expression as \(\binom{11}{3}\), so the result is also 165.

Analyzing the Options

Let's look at the given options:

• Option 1: 70
• Option 2: 165
• Option 3: \(^8 \text{C}_4\) - This represents choosing 4 items from 8, which is \(\frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70\). This would be relevant if we were choosing 4 distinguishable balls from 8 or arranging 8 items with 4 distinct types. It's not appropriate for this problem.
• Option 4: \(^8 \text{P}_4\) - This represents the number of permutations of 4 items chosen from 8, which is \(\frac{8!}{(8-4)!} = \frac{8!}{4!} = 8 \times 7 \times 6 \times 5 = 1680\). This is used when order matters and items are distinguishable. It's not appropriate for this problem.

Our calculated value is 165, which matches Option 2.

Conclusion on Placing 8 Indistinguishable Balls

Based on the stars and bars method, which is the correct approach for placing indistinguishable balls into distinguishable bins, there are 165 ways to distribute the 8 indistinguishable balls into the four distinguishable bins.

Revision Table: Combinations and Distributions

Additional Information: Stars and Bars Explained

The stars and bars method is a visual way to understand combinations with repetition. When you distribute \(n\) indistinguishable items into \(k\) distinguishable bins, it's equivalent to finding the number of non-negative integer solutions to the equation:

\[ x_1 + x_2 + \dots + x_k = n \]

where \(x_i\) is the number of items in bin \(i\). Each \(x_i\) must be greater than or equal to 0. The stars and bars method represents the \(n\) items as stars and uses \(k-1\) bars to divide them into \(k\) groups. The number of solutions is the number of ways to arrange these \(n\) stars and \(k-1\) bars, which is \(\binom{n+(k-1)}{k-1}\).

In our problem, \(n=8\) (balls) and \(k=4\) (bins), so we are looking for the number of non-negative integer solutions to \(x_1 + x_2 + x_3 + x_4 = 8\). The formula gives \(\binom{8+4-1}{4-1} = \binom{11}{3} = 165\).