If Quantity A is the number of ways to assign a number from 1 to 5 without repetition to each of four people, and Quantity B is the number of ways to assign a number from 1 to 5 without repetition to each of 5 people, then which of the following statements is correct with respect to Quantities A and B?
This question involves assigning unique items (numbers from 1 to 5) to a set of distinct positions (people) without repeating the items. This type of problem is addressed using permutations, as the order in which the numbers are assigned to different people matters.
The formula for calculating the number of permutations of selecting and arranging \(k\) items from a set of \(n\) distinct items is given by: $$P(n, k) = \frac{n!}{(n-k)!}$$ where \(n!\) (n factorial) is the product of all positive integers up to \(n\).
Quantity A is defined as the number of ways to assign a number from 1 to 5 without repetition to each of four people.
Using the permutation formula \(P(n, k)\):
$$ \text{Quantity A} = P(5, 4) = \frac{5!}{(5-4)!} $$ $$ P(5, 4) = \frac{5!}{1!} $$ $$ P(5, 4) = \frac{5 \times 4 \times 3 \times 2 \times 1}{1} $$ $$ P(5, 4) = \frac{120}{1} $$ $$ \text{Quantity A} = 120 $$So, there are 120 ways to assign numbers from 1 to 5 without repetition to four people.
Quantity B is defined as the number of ways to assign a number from 1 to 5 without repetition to each of 5 people.
Using the permutation formula \(P(n, k)\):
$$ \text{Quantity B} = P(5, 5) = \frac{5!}{(5-5)!} $$ $$ P(5, 5) = \frac{5!}{0!} $$By definition, \(0! = 1\).
$$ P(5, 5) = \frac{5!}{1} $$ $$ P(5, 5) = 5 \times 4 \times 3 \times 2 \times 1 $$ $$ \text{Quantity B} = 120 $$So, there are 120 ways to assign numbers from 1 to 5 without repetition to five people.
Now we compare the calculated values for Quantity A and Quantity B:
Since 120 is equal to 120, Quantity A and Quantity B are equal.
| Quantity | Description | Calculation | Value |
|---|---|---|---|
| Quantity A | Assign 1-5 without repetition to 4 people | \(P(5, 4) = \frac{5!}{(5-4)!} = 120\) | 120 |
| Quantity B | Assign 1-5 without repetition to 5 people | \(P(5, 5) = \frac{5!}{(5-5)!} = 120\) | 120 |
Based on the calculations, both quantities are equal.
| Concept | Description | Formula |
|---|---|---|
| Permutation | Number of ways to arrange \(k\) items from a set of \(n\) items where order matters and without repetition. | \(P(n, k) = \frac{n!}{(n-k)!}\) |
| Factorial | Product of all positive integers up to a given integer \(n\). | \(n! = n \times (n-1) \times \dots \times 2 \times 1\) |
| Zero Factorial | Defined value used in permutation/combination formulas. | \(0! = 1\) |
It's important to distinguish between permutations and combinations when solving counting problems.
In this question, since distinct numbers are assigned to distinct people, and the assignment to each person is unique, it is a permutation problem.
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