There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected

Q: Wait, why can't we just pick 4 men from the total pool of 9 men and 4 women from the total pool of 9 women?

That is a common trap! If you do that, you ignore the constraint that we must select exactly 4 people from Group A and 4 from Group B. The groups are distinct, and the composition of each group matters. You must respect the group-specific constraints while satisfying the total gender requirement.

Q: Does the order of selection matter here?

No, it does not. This is a selection problem, which means we are forming a team or a group. In such cases, the order in which you pick the individuals does not change the composition of the team. Therefore, we use combinations, not permutations.

Q: Why do we multiply the ways for Group A and Group B within each case?

This is the Fundamental Principle of Counting (Multiplication Rule). Since the selection from Group A and the selection from Group B are independent events that must both occur to form the final team, we multiply the number of ways to perform each selection.

Question

There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is _____

Sigma Percentile · Accepted Answer

### The Art of Combinatorial Balance

Welcome, future engineers! Today, we are going to unravel a problem that sits at the very heart of combinatorics. It is not just about crunching numbers; it is about organizing chaos.

Imagine you are standing in front of two rooms, Group A and Group B. In Group A, we have 4 men and 5 women. In Group B, we have 5 men and 4 women.

Our mission is to form a team of 8 people by selecting exactly 4 from each room, such that the final team has exactly 4 men and 4 women. This is a classic JEE problem that tests your ability to handle constraints systematically.

### The Constraint as an Anchor

When you face a problem with multiple constraints, the secret is to find an anchor. Here, our anchor is the number of men.

Let $m_A$ be the number of men we select from Group A, and $m_B$ be the number of men we select from Group B. We know that the total number of men must be 4, so we have the equation:

$$m_A + m_B = 4$$

Because we are selecting exactly 4 people from each group, once we fix $m_A$, the number of women from Group A is automatically fixed as $4 - m_A$. Similarly, $m_B$ determines the number of women from Group B as $4 - m_B$. This dependency is the key to unlocking the problem.

### Breaking Down the Cases

Since the cases are mutually exclusive, we must analyze each scenario separately. Let us walk through them:

**Case 1: $m_A = 4, m_B = 0$**

We select 4 men from Group A and 0 men from Group B. The number of ways is:

$$\binom{4}{4} 	imes \binom{5}{0} 	imes \binom{5}{0} 	imes \binom{4}{4} = 1 	imes 1 	imes 1 	imes 1 = 1 	ext{ way}$$

**Case 2: $m_A = 3, m_B = 1$**

Here, we select 3 men and 1 woman from Group A, and 1 man and 3 women from Group B. The calculation is:

$$\left( \binom{4}{3} 	imes \binom{5}{1} ight) 	imes \left( \binom{5}{1} 	imes \binom{4}{3} ight) = (4 	imes 5) 	imes (5 	imes 4) = 400 	ext{ ways}$$

**Case 3: $m_A = 2, m_B = 2$**

This is the most balanced case. We select 2 men and 2 women from each group. The calculation is:

$$\left( \binom{4}{2} 	imes \binom{5}{2} ight) 	imes \left( \binom{5}{2} 	imes \binom{4}{2} ight) = (6 	imes 10) 	imes (10 	imes 6) = 3600 	ext{ ways}$$

**Case 4: $m_A = 1, m_B = 3$**

We select 1 man and 3 women from Group A, and 3 men and 1 woman from Group B. The calculation is:

$$\left( \binom{4}{1} 	imes \binom{5}{3} ight) 	imes \left( \binom{5}{3} 	imes \binom{4}{1} ight) = (4 	imes 10) 	imes (10 	imes 4) = 1600 	ext{ ways}$$

**Case 5: $m_A = 0, m_B = 4$**

Finally, we select 0 men and 4 women from Group A, and 4 men and 0 women from Group B. The calculation is:

$$\left( \binom{4}{0} 	imes \binom{5}{4} ight) 	imes \left( \binom{5}{4} 	imes \binom{4}{0} ight) = (1 	imes 5) 	imes (5 	imes 1) = 25 	ext{ ways}$$

### The Final Summation

Now, we bring it all together. Since these cases are mutually exclusive, we apply the addition rule.

The total number of ways is the sum of all these cases:

$$1 + 400 + 3600 + 1600 + 25 = 5626$$

It is a beautiful, symmetric result. The lesson here is simple: when the problem feels overwhelming, break it down into smaller, manageable pieces. You have the tools; now go forth and solve with confidence! The final answer is **5626**.

Animated Solution for Mathematics - Permutations and Combinations: There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is _____

Enter Numerical Value:

Visualized Solution

Analyzing the Groups

Selection Constraints

Defining the Possible Cases

Case 1: $4 M$ from A, $0 M$ from B

Case 2: $3 M$ from A, $1 M$ from B

Case 3: $2 M$ from A, $2 M$ from B

Case 4: $1 M$ from A, $3 M$ from B

Case 5: $0 M$ from A, $4 M$ from B

Summing All Cases

Final Conclusion

The Sigma Insight: Combinations and Selection

The Art of Combinatorial Balance

The Constraint as an Anchor

Breaking Down the Cases

The Final Summation

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