A lock has a 3-number code made up of 29 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:

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Calculate Factorial of n: We need to calculate the number of ways to choose 3 different numbers from 29 without repetition. This is a permutation problem because the order of the numbers matters for the lock code. The formula for permutations of n items taken r at a time is nPr = n! / (n-r)!. Here, n = 29 and r = 3.Calculate Factorial of (n-r): First, calculate the factorial of n, which is 29! (29 factorial). However, we do not need to calculate the entire factorial value for 29! because it will be simplified in the permutation formula.Apply Permutation Formula: Next, calculate the factorial of (n-r), which is (29-3)! or 26!. Again, we do not need to calculate the entire factorial value for 26! because it will be simplified in the permutation formula.Perform Multiplication: Now, apply the permutation formula: 29P3 = 29! / (29-3)! = 29! / 26!. This simplifies to 29 × 28 × 27 because the 26! in the numerator and the denominator cancel each other out.Final Answer: Perform the multiplication: 29 × 28 × 27. 29 × 28 = 812, 812 × 27 = 21,924.The final answer is 21,924 different ways to choose three different numbers for a unique code.

A lock has a $3$ -number code made up of $29$ 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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A lock has a $3$ -number code made up of $29$ 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: