7 relatives of a man comprises 4 ladies and 3 gentlemen; his wife has also 7 relatives; 3 of them ar

Q: Wait, why can't we just pick 3 ladies from the total pool of 7 ladies and 3 gentlemen from the total pool of 7 gentlemen?

That is a classic trap! If you do that, you ignore the constraint that exactly 3 guests must come from the man's side and 3 from the wife's side. By picking from the total pool, you might end up with 5 guests from the man's side and 1 from the wife's side, which violates the problem's rules. We must respect the '3 from each side' constraint first.

Q: Does the order of selection matter here?

No, it does not. In a dinner party, the order in which you invite people doesn't change the composition of the group. That is why we use combinations ($\binom{n}{r}$) rather than permutations. We are simply forming a set of guests.

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

We add them because these cases are mutually exclusive. You cannot simultaneously have '3 ladies from the man's side' and '2 ladies from the man's side' in the same dinner party. Since these are distinct, non-overlapping scenarios that all satisfy the condition, the Fundamental Principle of Counting tells us to add the number of ways for each case.

Question

7 relatives of a man comprises 4 ladies and 3 gentlemen; his wife has also 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man's relatives and 3 of the wife's relatives?

Sigma Percentile · Accepted Answer

### The Dinner Party Dilemma: A Combinatorial Journey

Imagine you are planning a dinner party. It is not just any party; it is a delicate balancing act of social dynamics. You have two distinct pools of potential guests: the man's relatives and the wife's relatives.

Your goal is to invite exactly six people, ensuring that the final guest list is perfectly balanced with three ladies and three gentlemen. There is a constraint: you must invite exactly three people from the man's side and exactly three from the wife's side.

### The Two Worlds

Let us first look at our resources. The man's side of the family offers a pool of $7$ relatives: $4$ ladies and $3$ gentlemen.

The wife's side also offers $7$ relatives, but with a different composition: $3$ ladies and $4$ gentlemen. We are essentially bridging two different worlds to create a single, harmonious group of six.

### The Constraint Matrix

We need a total of $3$ ladies and $3$ gentlemen. Let $x$ be the number of ladies we choose from the man's side.

Since we must choose $3$ guests from his side, the number of gentlemen from his side will be $3-x$. Now, look at the wife's side.

To reach our total of $3$ ladies, we must choose $3-x$ ladies from her side. Consequently, to complete her quota of $3$ guests, we must choose $3-(3-x) = x$ gentlemen from her side. This dependency is the key to the entire problem.

### The Four Scenarios

We can now break this down into four mutually exclusive cases based on the value of $x$ (the number of ladies from the man's side).

**Case 1: $x=3$ (All 3 ladies from the man's side)**

Here, we select $3$ ladies and $0$ gentlemen from the man's side, and $0$ ladies and $3$ gentlemen from the wife's side. The math is elegant:

$$Ways = \binom{4}{3} 	imes \binom{3}{0} 	imes \binom{3}{0} 	imes \binom{4}{3} = 4 	imes 1 	imes 1 	imes 4 = 16$$

**Case 2: $x=2$ (2 ladies, 1 gentleman from the man's side)**

Now we need $1$ lady and $2$ gentlemen from the wife's side. The calculation is as follows:

$$Ways = \binom{4}{2} 	imes \binom{3}{1} 	imes \binom{3}{1} 	imes \binom{4}{2} = 6 	imes 3 	imes 3 	imes 6 = 324$$

**Case 3: $x=1$ (1 lady, 2 gentlemen from the man's side)**

Here, we need $2$ ladies and $1$ gentleman from the wife's side. The calculation is as follows:

$$Ways = \binom{4}{1} 	imes \binom{3}{2} 	imes \binom{3}{2} 	imes \binom{4}{1} = 4 	imes 3 	imes 3 	imes 4 = 144$$

**Case 4: $x=0$ (0 ladies, 3 gentlemen from the man's side)**

Finally, we need $3$ ladies and $0$ gentlemen from the wife's side. The calculation is as follows:

$$Ways = \binom{4}{0} 	imes \binom{3}{3} 	imes \binom{3}{3} 	imes \binom{4}{0} = 1 	imes 1 	imes 1 	imes 1 = 1$$

### The Grand Summation

Because these four scenarios are distinct, we simply add them together to find the total number of ways to invite our guests.

$$Total = 16 + 324 + 144 + 1 = 485$$

And there you have it! By breaking a complex constraint into manageable, logical pieces, we have navigated the social complexities of this dinner party.

Remember, in combinatorics, the secret is often not in the calculation itself, but in how you structure the problem. The final answer is **485**.

Enter Numerical Value:

Visualized Solution

Man's Relatives

Wife's Relatives

Selection Constraints

Strategy for Cases

Case 1: 3 Ladies from Man's Side

Case 2: 2 Ladies from Man's Side

Case 3: 1 Lady from Man's Side

Case 4: 0 Ladies from Man's Side

Total Number of Ways

The Sigma Insight: Combinations and Selection

The Dinner Party Dilemma

The Two Worlds

The Constraint Matrix

The Four Scenarios

The Grand Summation

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