Let S = {1, 2, 3, ....., 9}. For k = 1, 2, ....., 5, let Nₖ be the number of subsets of S, each cont

Q: Wait, why can't I just calculate $N_1, N_2, N_3, N_4, N_5$ separately? It seems safer.

Ah, the siren song of brute force! While it is technically possible, it is a dangerous path in the JEE. You have five separate calculations to perform, each with its own chance for a silly arithmetic error. In competitive exams, time is your most precious currency. Always pause to see if the sum of cases represents a simpler, total set.

Q: Wait, why is $N_0$ impossible? I don't see it.

Look at the constraints of your set. You need five elements for your subset. If you want zero odd numbers, you must pick all five from the even numbers. But count them again: $\{2, 4, 6, 8\}$. There are only four! You cannot pick five items from a group of four. It is physically impossible, so $N_0 = 0$.

Q: Is this logic applicable to other problems?

Absolutely! This is the 'Complementary Counting' or 'Total Case Analysis' mindset. Whenever you see a sum of cases, ask yourself: 'Does this sum cover the entire sample space?' If it does, you can often bypass the individual calculations entirely.

Question

Let $S = \{1, 2, 3, ....., 9\}$. For $k = 1, 2, ....., 5$, let $N_k$ be the number of subsets of S, each containing five elements out of which exactly $k$ are odd. Then $N_1 + N_2 + N_3 + N_4 + N_5 =$

Options:

(A) 210

(B) 252

(C) 125

(D) 126

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are working with the set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. This set contains nine distinct elements, partitioned into five odd numbers $\{1, 3, 5, 7, 9\}$ and four even numbers $\{2, 4, 6, 8\}$.

Our goal is to determine the sum $N_1 + N_2 + N_3 + N_4 + N_5$, where $N_k$ represents the number of subsets of size five containing exactly $k$ odd numbers.

### The Trap of Brute Force

A common instinct is to calculate each $N_k$ individually by considering the number of ways to choose $k$ odd numbers and $(5-k)$ even numbers. However, calculating five separate combinations is inefficient and prone to arithmetic errors.

In the context of competitive examinations like the JEE, time management is critical. We must look for a structural shortcut rather than performing exhaustive calculations.

### The Grand Realization

The total number of ways to choose five elements from nine is given by the binomial coefficient $\binom{9}{5}$. This total must equal the sum of all possible cases for the number of odd elements:

$$N_0 + N_1 + N_2 + N_3 + N_4 + N_5 = \binom{9}{5}$$

Consider $N_0$, which represents the number of subsets with zero odd elements. This would require choosing all five elements from the set of even numbers.

Since there are only four even numbers available in $S$, it is impossible to select five even numbers. Therefore, we conclude that $N_0 = 0$.

### The Elegant Conclusion

Because $N_0 = 0$, the sum we seek simplifies significantly:

$$N_1 + N_2 + N_3 + N_4 + N_5 = \binom{9}{5}$$

Using the property of symmetry in binomial coefficients, we know that $\binom{9}{5} = \binom{9}{4}$. We calculate this value as follows:

$$\binom{9}{4} = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1}$$

Simplifying the expression:

$$\binom{9}{4} = 9 	imes 7 	imes 2 = 126$$

The final result is **126**. By identifying the underlying structure of the problem, we have bypassed the need for complex summations and arrived at the solution with absolute precision.

Animated Solution for Mathematics - Permutations and Combinations: Let $S = {1, 2, 3, ....., 9}$ . For $k = 1, 2, ....., 5$ , let $N_{k}$ be the number of subsets of S, each containing five elements out of which exactly $k$ are odd. Then $N_{1} + N_{2} + N_{3} + N_{4} + N_{5} =$

Select Answer:

Visualized Solution

Analyze the Set $S$

Categorize Elements

Understand $N_{k}$

The Long Method (Formula)

The JEE Trap

The Smart Approach

The Impossible Case

The Grand Realization

Final Calculation

Compute the Value

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Trap of Brute Force

The Grand Realization

The Elegant Conclusion

Similar Questions

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Select Answer:

Visualized Solution

Analyze the Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Categorize Elements

Understand .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Nk​

The Long Method (Formula)

The JEE Trap

The Smart Approach

The Impossible Case

The Grand Realization

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The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Trap of Brute Force

The Grand Realization

The Elegant Conclusion

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