Answer :
The correct answer is 120.
Given that Sharon’s ZIP Code contains the digits 7, 3, 6, 1, and 2, we need to arrange these digits to form a 5-digit code. Since no repetition of digits is allowed (each digit can appear only once), we’ll use the concept of permutations.
The total number of ways to arrange these 5 digits is given by:
[tex][ \text{Number of arrangements} = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 ][/tex]
Therefore, there are 120 different possibilities for Sharon’s ZIP Code.
The correct answer is 120.
Answer:
B
Step-by-step explanation:
There are 5 spots to fill with the digits 7, 3, 6, 1, and 2 to make a code
Think how many ways you can fill each of the spots.
spot 1: either 7, 3, 6, 1, or 2 so there are 5 ways to fill that first spot
spot 2: is can be filled in 4 ways because is given that each of the digits have to be in the code so if we used one of the numbers for the first spot now we have 4 numbers left to use
spot 3: 3 ways , spot2: 2 ways, and spot1: 1 way
The different possibilities for Sharon's ZIP Code are given by the multiplication of the numbers of ways you can fill each spot in the code 5*4*3*2*1 = 120
