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A lock has a 3-number code made up of 21 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 $21$ 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

21

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 combinations possible for a $3$ -number code from $21$ unique numbers without repetition. This is a permutation problem because the order of the numbers matters for the code. $\newline$ To calculate the number of permutations of $n$ items taken $r$ at a time, we use the formula: $\newline$ $P(n, r) = \frac{n!}{(n - r)!}$ $\newline$ In this case, $n = 21$ (total numbers) and $r = 3$ (numbers in the code).
Apply Permutation Formula: First, we calculate the factorial of $21$ ( $21!$ ): $21! = 21 \times 20 \times 19 \times \ldots \times 1$ However, we don't need to calculate the entire factorial because the denominator will cancel out most of the terms.
Perform Multiplication: Next, we calculate the factorial of $(21 - 3)$ which is $18$ $(18!)$ : $18! = 18 \times 17 \times 16 \times \ldots \times 1$ Again, we don't need to calculate the entire factorial for the same reason as above.
Perform Multiplication: Next, we calculate the factorial of $(21 - 3)$ which is $18$ $(18!)$ : $18! = 18 \times 17 \times 16 \times \ldots \times 1$ Again, we don't need to calculate the entire factorial for the same reason as above.Now, we apply the permutation formula: $P(21, 3) = \frac{21!}{(21 - 3)!}$ $= \frac{21!}{18!}$ $= \frac{21 \times 20 \times 19}{(18 \times 17 \times 16 \times \ldots \times 1)}$ The terms from $18!$ in the denominator will cancel out the corresponding terms in the numerator, leaving us with: $= 21 \times 20 \times 19$
Perform Multiplication: Next, we calculate the factorial of $(21 - 3)$ which is $18$ $(18!)$ : $18! = 18 \times 17 \times 16 \times \ldots \times 1$ Again, we don't need to calculate the entire factorial for the same reason as above.Now, we apply the permutation formula: $P(21, 3) = \frac{21!}{(21 - 3)!}$ $= \frac{21!}{18!}$ $= \frac{21 \times 20 \times 19}{(18 \times 17 \times 16 \times \ldots \times 1)}$ The terms from $18!$ in the denominator will cancel out the corresponding terms in the numerator, leaving us with: $= 21 \times 20 \times 19$ We perform the multiplication: $21 \times 20 \times 19 = 7980$ So, there are $7980$ different ways to choose three different numbers for the lock code.

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