Math Snap

STEP 1

1. ${}^{n}P_{3}$ represents the number of permutations of 3 items from a set of $n$ items.
2. ${}^{n}C_{5}$ represents the number of combinations of 5 items from a set of $n$ items.
3. The equation ${}^{n}P_{3} = 4 \times {}^{n}C_{5}$ needs to be solved for $n$.

STEP 2

1. Express the permutation and combination formulas.
2. Substitute these expressions into the given equation.
3. Simplify the equation and solve for $n$.

STEP 3

Write the formula for permutations and combinations:
The formula for permutations is:
$${}^{n}P_{r} = \frac{n!}{(n-r)!}$$ The formula for combinations is:
$${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$$ For this problem, we have:
$${}^{n}P_{3} = \frac{n!}{(n-3)!}$$ $${}^{n}C_{5} = \frac{n!}{5!(n-5)!}$$

STEP 4

Substitute these expressions into the given equation:
$$\frac{n!}{(n-3)!} = 4 \times \frac{n!}{5!(n-5)!}$$

STEP 5

Cancel $n!$ from both sides of the equation:
$$\frac{1}{(n-3)!} = \frac{4}{5!(n-5)!}$$

STEP 6

Cross-multiply to eliminate the fractions:
$$5!(n-5)! = 4(n-3)!$$

STEP 7

Recognize that $(n-3)! = (n-3)(n-4)(n-5)!$. Substitute this into the equation:
$$5!(n-5)! = 4(n-3)(n-4)(n-5)!$$ Cancel $(n-5)!$ from both sides:
$$5! = 4(n-3)(n-4)$$

STEP 8

Calculate $5!$ and solve for $n$:
$$5! = 120$$ $$120 = 4(n-3)(n-4)$$ Divide both sides by 4:
$$30 = (n-3)(n-4)$$ Expand the right side:
$$30 = n^2 - 7n + 12$$ Rearrange to form a quadratic equation:
$$n^2 - 7n + 12 - 30 = 0$$ $$n^2 - 7n - 18 = 0$$

Factor the quadratic equation:
$$(n - 9)(n + 2) = 0$$ Set each factor to zero and solve for $n$:
$$n - 9 = 0 \quad \Rightarrow \quad n = 9$$ $$n + 2 = 0 \quad \Rightarrow \quad n = -2$$ Since $n$ must be a positive integer (as it represents a count of items), we choose:
$$n = 9$$ The value of $n$ is:
$$\boxed{9}$$