Suppose Anil's mother wants to give 5 whole fruits to Anil from a basket of 7 red apples, 5 white ap

Q: Why can't we use the 'stars and bars' method here?

Stars and bars is perfect for distributing identical items into distinct bins, but here, we are selecting items from a finite pool of distinct types. Because the number of available fruits (8, 7, 5) is limited, we must use combinations (${}^nC_r$) to respect those limits.

Q: How do I know I've found all the cases?

Great question! The best way is to be systematic. Start by fixing the minimums: $O=2, R=1, W=1$. That uses up 4 fruits. You have 1 fruit left to distribute. You can give it to $O$, $R$, or $W$. That gives you exactly 3 scenarios. If you try to add more, you'll exceed the total of 5.

Q: Why do we multiply the combinations within a case?

This is the Fundamental Counting Principle. Since the selection of oranges, red apples, and white apples are independent events that must *all* happen to complete the selection of 5 fruits, we multiply the number of ways for each event.

Question

Suppose Anil's mother wants to give 5 whole fruits to Anil from a basket of 7 red apples, 5 white apples and 8 oranges. If in the selected 5 fruits, at least 2 orange, at least one red apple and at least one white apple must be given, then the number of ways, Anil's mother can offer 5 fruits to Anil is ____.

Sigma Percentile · Accepted Answer

### The Art of the Fruit Basket: A Combinatorial Journey

Imagine you are standing in front of a bountiful basket containing 8 oranges, 7 red apples, and 5 white apples. Your task is to select exactly 5 fruits subject to specific constraints: at least 2 oranges, at least 1 red apple, and at least 1 white apple.

This is a puzzle of constraints. Let us break it down systematically.

### Analyzing the Setup

We are selecting a total of $n = 5$ fruits. Let $O$ be the number of oranges, $R$ be the number of red apples, and $W$ be the number of white apples.

Our constraints are defined as:
$O \ge 2$, $R \ge 1$, and $W \ge 1$.

Given the total sum $O + R + W = 5$, we identify the possible integer combinations that satisfy these inequalities. Starting with the minimum requirements ($2+1+1=4$), we have exactly 1 fruit remaining to distribute.

This leads to three distinct, mutually exclusive cases:
1. Case 1: $(O=2, R=1, W=2)$
2. Case 2: $(O=2, R=2, W=1)$
3. Case 3: $(O=3, R=1, W=1)$

### The Three Paths

We calculate the number of ways for each path using the combination formula:
$$^{n}C_{r} = \frac{n!}{r!(n-r)!}$$

**For Case 1 ($2$ Oranges, $1$ Red Apple, $2$ White Apples):**
The number of ways is calculated as:
$$^{8}C_{2} 	imes ^{7}C_{1} 	imes ^{5}C_{2} = 28 	imes 7 	imes 10 = 1960$$

**For Case 2 ($2$ Oranges, $2$ Red Apples, $1$ White Apple):**
The number of ways is calculated as:
$$^{8}C_{2} 	imes ^{7}C_{2} 	imes ^{5}C_{1} = 28 	imes 21 	imes 5 = 2940$$

**For Case 3 ($3$ Oranges, $1$ Red Apple, $1$ White Apple):**
The number of ways is calculated as:
$$^{8}C_{3} 	imes ^{7}C_{1} 	imes ^{5}C_{1} = 56 	imes 7 	imes 5 = 1960$$

### Final Calculation

To find the total number of ways to select the fruits, we sum the results of the three mutually exclusive cases:
$$	ext{Total Ways} = 1960 + 2940 + 1960$$

$$	ext{Total Ways} = \mathbf{6860}$$

By breaking the complex constraints into simple, manageable cases, we have determined that there are exactly **6860** ways to select the fruits.

Enter Numerical Value:

Visualized Solution

Understanding the Inventory

Identifying Possible Combinations

Case 1: $(2 O, 1 R, 2 W)$ Setup

Case 1: Calculation

Case 2: $(2 O, 2 R, 1 W)$ Setup

Case 2: Calculation

Case 3: $(3 O, 1 R, 1 W)$ Setup

Case 3: Calculation

Summing All Possibilities

Final Result

The Sigma Insight: Combinations and Selection

The Art of the Fruit Basket

Analyzing the Setup

The Three Paths

Final Calculation

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