Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of

Q: Wait, why can't we just pick 4 boys from the total pool of 13 boys and 4 girls from the total pool of 8 girls?

That is a classic trap! The problem explicitly states that 5 people must come from Group A and 3 from Group B. If you just pick from the total pool, you ignore the group-specific constraints. We must respect the '5 from A, 3 from B' rule, which forces us to break the problem into cases based on how many boys and girls are taken from each group.

Q: How do we know $y$ (the number of girls from Group A) can only be 1, 2, or 3?

Great observation! We know $x+y=5$ (total from Group A) and $x+u=4$ (total boys). Since $x$ is the number of boys from Group A, $x$ cannot exceed the total number of boys needed (4). If $x \le 4$, then $y = 5-x$ must be $\ge 1$. Also, Group A only has 3 girls, so $y$ cannot be more than 3. Thus, $y$ must be 1, 2, or 3.

Q: Why do we add the results of the cases instead of multiplying them?

We add them because these cases are mutually exclusive. You either pick 1 girl from Group A (Case 1) OR 2 girls from Group A (Case 2) OR 3 girls from Group A (Case 3). You cannot do both simultaneously. In combinatorics, when you have distinct, non-overlapping scenarios that satisfy the same goal, you sum the number of ways for each scenario.

Question

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to:

Options:

(A) 8750

(B) 9100

(C) 8925

(D) 8575

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a room with two distinct groups of students. Group A is a vibrant mix of $7$ boys and $3$ girls, while Group B holds $6$ boys and $5$ girls.

Your mission is to curate a picnic party of exactly $8$ people—$4$ boys and $4$ girls—subject to a very specific, rigid constraint: you must select exactly $5$ people from Group A and exactly $3$ from Group B.

Let $x$ and $y$ be the number of boys and girls selected from Group A, respectively. Since we must pick $5$ people from Group A, we have the equation:
$$x + y = 5$$

Similarly, let $u$ and $v$ be the number of boys and girls from Group B. Since we must pick $3$ people from Group B, we have:
$$u + v = 3$$

Our ultimate goal is to satisfy the picnic requirement for the total number of boys and girls:
$$x + u = 4 \quad 	ext{and} \quad y + v = 4$$

### The Constraint Web

We need to find the possible values for $y$, the number of girls from Group A. Since the total number of boys required is $4$, the number of boys from Group A, $x$, cannot exceed $4$.

If $x \le 4$, then from our first equation $x + y = 5$, it follows that $y$ must be at least $1$. Furthermore, Group A only contains $3$ girls, so $y$ cannot exceed $3$.

Thus, our possible values for $y$ are $1, 2,$ and $3$. This gives us three distinct, mutually exclusive cases to analyze.

### Case-by-Case Analysis

**Case 1: $y = 1$**

If we select $1$ girl from Group A, we must select $x = 5 - 1 = 4$ boys from Group A. To reach our total of $4$ boys, we need $u = 4 - 4 = 0$ boys from Group B. To complete our selection of $3$ people from Group B, we need $v = 3 - 0 = 3$ girls.

The number of ways is:
$$\binom{7}{4} 	imes \binom{3}{1} 	imes \binom{6}{0} 	imes \binom{5}{3} = 35 	imes 3 	imes 1 	imes 10 = 1050$$

**Case 2: $y = 2$**

If we select $2$ girls from Group A, we need $x = 5 - 2 = 3$ boys from Group A. To reach our total of $4$ boys, we need $u = 4 - 3 = 1$ boy from Group B. To complete our selection of $3$ people from Group B, we need $v = 3 - 1 = 2$ girls.

The number of ways is:
$$\binom{7}{3} 	imes \binom{3}{2} 	imes \binom{6}{1} 	imes \binom{5}{2} = 35 	imes 3 	imes 6 	imes 10 = 6300$$

**Case 3: $y = 3$**

If we select $3$ girls from Group A, we need $x = 5 - 3 = 2$ boys from Group A. To reach our total of $4$ boys, we need $u = 4 - 2 = 2$ boys from Group B. To complete our selection of $3$ people from Group B, we need $v = 3 - 2 = 1$ girl.

The number of ways is:
$$\binom{7}{2} 	imes \binom{3}{3} 	imes \binom{6}{2} 	imes \binom{5}{1} = 21 	imes 1 	imes 15 	imes 5 = 1575$$

### The Grand Total

Because these cases are mutually exclusive, we simply sum them up:
$$1050 + 6300 + 1575 = 8925$$

There you have it! By breaking the problem down into manageable, logical constraints, we have navigated the complexity and arrived at the final answer of **$8925$ ways**.

Animated Solution for Mathematics - Permutations and Combinations: Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to:

Select Answer:

Visualized Solution

Understanding the Groups

The Picnic Requirement

Selection Constraints

Mathematical Setup

Total Boys and Girls

Finding Possible Cases

Case 1: $y = 1$

Calculating Case 1

Case 2: $y = 2$

Calculating Case 2

Case 3: $y = 3$

Calculating Case 3

Total Number of Ways

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Constraint Web

Case-by-Case Analysis

The Grand Total

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