Question

Jenna has the following cards: 3, 4, 5, 6, 7. Without repeating digits how many different three-digit numbers can be made using the cards

Answer 1

Answer:

There are 60 different 3-digit numbers.

Step-by-step explanation:

Arrangement of 3-digit numbers → order is significant → Permutation Probability

$\boxed{_nP_r=\frac{n!}{(n-r)!} }$

Given:

total no. of objects (n) = 5 (all of 5 are distinctive numbers)

total no. of selected objects (r) = 3

$\displaystyle _5P_3=\frac{5!}{(5-3)!}$

     $\displaystyle =\frac{5\times4\times3\times2\times1}{2\times1}$

     $=60$

Answer 2

Answer:

$60$.

Step-by-step explanation:

The question states that the cards are drawn without repeating digits, meaning that cards would not be replaced. The number of possible combinations would need to be found using combinatorics.

The next step is to determine if the ordering of the cards would make a difference. Since the cards are all different, the ordering of the cards would indeed make a difference. For example, the number $\texttt{567}$ would not be the same as $\texttt{765}$,.

Hence, the total number of possible combinations would be equivalent to the number of ways to select and order ("permute") $3$ of $5$ distinct items without replacement:

$\displaystyle P(5,\, 3) = \frac{5!}{(5 - 3)!} = 5\times 4 \times 3 = 60$.

In other words, these cards can form a total of $60$ possible three-digit numbers.