If no digit appears more than once, how many 2-digit numbers can be formed from the digits 1, 3, 5, and 7?
P(4,2) = 12
Thank you
To determine the number of 2-digit numbers that can be formed from the digits 1, 3, 5, and 7 without repetition, we need to count the number of possibilities.
Step 1: Determine the number of choices for the first digit.
Since the first digit cannot be zero, we have three options: 1, 3, and 5.
Step 2: Determine the number of choices for the second digit.
Since we cannot repeat any digit, we have three remaining options after choosing the first digit.
Step 3: Multiply the number of choices for each step.
To calculate the total number of possibilities, we multiply the number of choices for each step: 3 options for the first digit multiplied by 3 options for the second digit. So, the total number of 2-digit numbers that can be formed is 3 x 3 = 9.
Therefore, there are 9 different 2-digit numbers that can be formed from the digits 1, 3, 5, and 7 without repetition.
To find the number of 2-digit numbers that can be formed from the digits 1, 3, 5, and 7, where no digit appears more than once, we can use the concept of permutation.
A permutation is an arrangement of objects in a specific order. In this case, we are arranging the digits to form 2-digit numbers.
To find the number of permutations, we can use the formula:
nPr = n! / (n - r)!
Where n is the total number of objects (digits) available and r is the number of objects (digits) to be chosen.
In this case, we have 4 digits (1, 3, 5, and 7) to choose from and we need to form 2-digit numbers. So, n = 4 and r = 2.
Plugging these values into the formula, we get:
4P2 = 4! / (4 - 2)!
= 4! / 2!
= (4 x 3 x 2 x 1) / (2 x 1)
= 24 / 2
= 12
Therefore, there are 12 different 2-digit numbers that can be formed from the digits 1, 3, 5, and 7, where no digit appears more than once.