Question

Math

Posted 4 months ago

`X is a finite set with the following property: The number of subsets of X that contain exactly 3 elements is 14 more than the number of subsets of X that contain exactly 2 elements. How many subsets of X contain exactly 4 elements?`

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Answer from Sia

Posted 4 months ago

Solution by Steps

step 1

Given that the number of subsets of $X$ with exactly 3 elements is 14 more than the number of subsets with exactly 2 elements, we can write this as $\binom{n}{3} = \binom{n}{2} + 14$

step 2

We know that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, so we can express the given equation in factorial terms as $\frac{n!}{3!(n-3)!} = \frac{n!}{2!(n-2)!} + 14$

step 3

Simplifying the equation, we get $\frac{n(n-1)(n-2)}{6} = \frac{n(n-1)}{2} + 14$

step 4

Multiplying through by 6 to clear the denominators gives us $n(n-1)(n-2) = 3n(n-1) + 84$

step 5

Expanding and simplifying, we get $n^3 - 3n^2 + 2n = 3n^2 - 3n + 84$

step 6

Subtracting $3n^2 - 3n$ from both sides, we obtain $n^3 - 6n^2 + 5n - 84 = 0$

step 7

Factoring the cubic equation, we find that $n = 7$ is a solution

step 8

Now we need to find the number of subsets of $X$ with exactly 4 elements, which is $\binom{n}{4}$

step 9

Substituting $n = 7$ into $\binom{n}{4}$, we get $\binom{7}{4} = \frac{7!}{4!3!}$

step 10

Calculating the factorial values, we find $\binom{7}{4} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35$

Answer

There are 35 subsets of $X$ that contain exactly 4 elements.

Key Concept

Combinatorial identities and binomial coefficients

Explanation

The problem is solved by setting up an equation based on the given property of the set $X$ and solving for $n$, the number of elements in $X$. Then, using the binomial coefficient formula, we calculate the number of subsets containing exactly 4 elements.

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