# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-8/6

Exercises: Try Out - Set 8: Real exam simulation to sharpen your skills and build confidence. - Problem 6

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## Exercise 6

### Question

export const metadata = {
  title: "Problem 6",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Set <InlineMath math="A" /> has <InlineMath math="10" /> members. The number of subsets of <InlineMath math="A" /> that have more than <InlineMath math="3" /> members is...


### Choices

- [ ] $$512$$
- [ ] $$564$$
- [ ] $$624$$
- [ ] $$720$$
- [x] $$848$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 6",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

The order of members in a set does not matter, so we use the concept of combinations.

The total number of subsets of a set with <InlineMath math="n" /> members is <InlineMath math="2^n" />. Given <InlineMath math="n = 10" />, the total number of subsets is:

<BlockMath math="2^{10} = 1024" />

We are asked to find the number of subsets with **more than <InlineMath math="3" />** members. This means we are looking for the sum of subsets with 4, 5, 6, ..., 10 members.

An easier way is to use the complement principle: Total subsets minus subsets with **<InlineMath math="0" />, <InlineMath math="1" />, <InlineMath math="2" />, or <InlineMath math="3" />** members.

The combination formula is:

<BlockMath math="C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}" />

Let's calculate them one by one.

#### Subsets with 0 members

<BlockMath math="C(10, 0) = \frac{10!}{0!10!} = 1" />

#### Subsets with 1 member

<BlockMath math="C(10, 1) = \frac{10!}{1!9!} = 10" />

#### Subsets with 2 members

<BlockMath math="C(10, 2) = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 45" />

#### Subsets with 3 members

<BlockMath math="C(10, 3) = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120" />

The number of subsets with <InlineMath math="\leq 3" /> members is:

<BlockMath math="1 + 10 + 45 + 120 = 176" />

Thus, the number of subsets with **more than <InlineMath math="3" />** members is:

<BlockMath math="\begin{aligned} \text{Total} - (\leq 3) &= 1024 - 176 \\ &= 848 \end{aligned}" />

So, the number of subsets of <InlineMath math="A" /> that have more than <InlineMath math="3" /> members is <InlineMath math="848" />.


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