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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∻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.⚹◊What is the total number of subsets in set X?⍭ Generate me a similar question◊

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