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A lock has a 3-number code made up of 28 numbers. If none of the numbers are allowed to repeat, how many different ways can you choose three different numbers in order for a unique code?
Answer:

A lock has a $3$ -number code made up of $28$ numbers. If none of the numbers are allowed to repeat, how many different ways can you choose three different numbers in order for a unique code? $\newline$ Answer:

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Experimental probability

Full solution

Q. A lock has a

3

-number code made up of

28

numbers. If none of the numbers are allowed to repeat, how many different ways can you choose three different numbers in order for a unique code?

\newline

Answer:

Calculate Factorials: We need to calculate the number of ways to choose $3$ different numbers from $28$ without repetition. This is a permutation problem because the order of the numbers matters for the code. $\newline$ The formula for permutations of $n$ items taken $r$ at a time is $nPr = \frac{n!}{(n-r)!}$ . $\newline$ Here, $n = 28$ (total numbers) and $r = 3$ (numbers in the code).
Apply Permutation Formula: First, we calculate the factorial of $28$ , which is $28! = 28 \times 27 \times 26 \times \ldots \times 1$ . However, we do not need to calculate the entire factorial. We only need the product of the first three terms because the rest will cancel out when we divide by $(28-3)!$ .
Simplify Calculation: Now, we calculate the factorial of $(28-3)$ , which is $25! = 25 \times 24 \times 23 \times \ldots \times 1$ . Again, we do not need the entire factorial for $25!$ because it will cancel out with the $28!$ in the numerator.
Perform Multiplication: We can now write the permutation formula as $28P3 = \frac{28!}{(28-3)!} = \frac{28 \times 27 \times 26}{25 \times 24 \times 23 \times ... \times 1}$ . The terms from $25!$ in the denominator cancel out with the corresponding terms in the numerator after $28 \times 27 \times 26$ .
Final Answer: After canceling, we are left with $28P3 = 28 \times 27 \times 26$ . Now we perform the multiplication: $28 \times 27 \times 26 = 19656$ .
Final Answer: After canceling, we are left with $28P3 = 28 \times 27 \times 26$ . Now we perform the multiplication: $28 \times 27 \times 26 = 19656$ . The final answer is the number of different $3$ -number codes that can be made from $28$ numbers without repeating any numbers, which is $19656$ .

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