Let ∪_{i=1}^{50} X_i = ∪_{i=1}^{n} Y_i = T, where each X_i contains 10 elements and each Y_i contain

Q: Wait, why do we divide by 20? Shouldn't we subtract something?

That is a common confusion! We use division here because we are dealing with a 'frequency' or 'multiplicity' problem, not an inclusion-exclusion problem. If an element appears in 20 sets, it means it has been 'counted' 20 times in the total sum of the sizes of the sets. To get back to the unique count, we simply divide the total sum by that frequency. Think of it like this: if you have 20 copies of a book, you don't have 20 different books; you have 1 book counted 20 times.

Q: Does the order of the sets $X_i$ or $Y_i$ matter?

Not at all! In set theory, the order of sets in a collection does not change the union. Whether you label them $X_1, X_2, \dots, X_{50}$ or in any other order, the total number of unique elements in their union remains the same. The problem relies only on the total count and the frequency of each element.

Q: What if an element appeared in a different number of sets?

That would make the problem much more complex! If the frequency varied, we would need to know the specific distribution of elements across the sets. However, in this problem, the phrasing 'each element of the set $T$ is an element of exactly 20 of sets $X_i$'s' guarantees a uniform frequency, which is the key that allows us to use this elegant division method.

Question

Let $\cup_{i=1}^{50} X_i = \cup_{i=1}^{n} Y_i = T$, where each $X_i$ contains 10 elements and each $Y_i$ contains 5 elements. If each element of the set $T$ is an element of exactly 20 of sets $X_i$'s and exactly 6 of sets $Y_i$'s then $n$ is equal to

Options:

(A) 50

(B) 15

(C) 30

(D) 45

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a massive, shimmering container, which we will call set $T$. This container is a masterpiece constructed by two different groups of architects.

On one side, we have a group of 50 sets, $X_1, X_2, \dots, X_{50}$. On the other, we have a group of $n$ sets, $Y_1, Y_2, \dots, Y_n$. Both groups, when combined, perfectly fill our container $T$.

### Phase 1: The Naive Count and the Correction

Let's start with the first group of architects, the $X$ collection. We are told there are 50 sets, and each set $X_i$ contains exactly 10 elements.

If we were to walk up to each of these 50 sets and write down every element we see, we would have a total of 500 entries:
$$50 	imes 10 = 500$$

However, this is a 'naive' count that assumes every element is unique. The problem states that every single element in our container $T$ is actually present in exactly 20 of these $X$ sets.

To find the true, unique number of elements in $T$, we must divide our total count by this repetition factor:
$$|T| = \frac{500}{20} = 25$$

The fog is lifting; our container $T$ holds exactly 25 unique elements.

### Phase 2: The Bridge to the Y-Collection

Now, let's turn our attention to the second group of architects, the $Y$ collection. We have $n$ sets, and each set $Y_i$ contains 5 elements.

If we perform the same naive count, we get a total of $5n$ entries. We are told that each element of $T$ is present in exactly 6 of these $Y$ sets.

To find the true size of $T$ using this second perspective, we apply the same logic:
$$|T| = \frac{5n}{6}$$

### Phase 3: The Grand Synthesis

We have now calculated the size of the exact same container $T$ in two different ways. Since these two expressions describe the same set, they must be equal:
$$\frac{5n}{6} = 25$$

Now, we solve for $n$. Multiply both sides by 6 to clear the fraction:
$$5n = 25 	imes 6$$
$$5n = 150$$

Finally, divide by 5 to isolate $n$:
$$n = \frac{150}{5} = 30$$

And there we have it! The number of sets in the second collection, **$n = 30$**.

We didn't need to know which elements were where; we only needed to understand the relationship between the total count and the frequency of occurrence. Keep this **'overcounting' principle** in your toolkit—it is a powerful weapon for any JEE aspirant!

Select Answer:

Visualized Solution

The Universal Set $T$

The Collection of Sets $X_{i}$

Total Elements Before Removing Duplicates

The Overcounting Factor for $X_{i}$

Calculating the True Size of $T$

The Collection of Sets $Y_{i}$

Total Elements in $Y_{i}$ Before Removing Duplicates

The Overcounting Factor for $Y_{i}$

Expressing $∣ T ∣$ using $Y_{i}$

Equating the Two Values of $∣ T ∣$

Solving for $n$

Final Answer

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Visualized Solution

The Universal Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

The Collection of Sets .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Xi​

Total Elements Before Removing Duplicates

The Overcounting Factor for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Xi​

Calculating the True Size of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

The Collection of Sets .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Yi​

Total Elements in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Yi​ Before Removing Duplicates

The Overcounting Factor for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Yi​

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣T∣ using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Yi​

Equating the Two Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣T∣

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n