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Please help me :/

You can make a 6-digit security number using the digits 1-9 and digits cannot be repeated. Show all work and formulas used in computing your answers.



a) How many numbers can you make if there are no additional restrictions?



b) How many numbers can you make if the first digit cannot be a one?



c) How many odd numbers can you make (the last digit is odd?)



d) How many numbers greater than 300,000 can you make?



e) How many numbers greater than 750,000 can you make?

asked
User Mniess
by
8.2k points

1 Answer

3 votes

Sure, I'd be happy to help you with these questions!

a) To calculate the total number of possible 6-digit security numbers, we can use the permutation formula:

nPr = n! / (n-r)!

where n is the total number of digits available (9) and r is the number of digits we are selecting (6).

So, the number of possible 6-digit security numbers without any restrictions is:

9P6 = 9! / (9-6)! = 9! / 3! = 9 x 8 x 7 x 6 x 5 x 4 = 60,480

Therefore, there are 60,480 possible 6-digit security numbers that can be made with the digits 1-9 without repeating any digits.

b) If the first digit cannot be a one, we are left with 8 choices for the first digit (since we cannot use 1) and 8 choices for the second digit (since we have already used one digit). For the remaining 4 digits, we still have 7 choices for each digit, since we cannot repeat any digits.

Using the permutation formula again, the number of possible 6-digit security numbers with the first digit not being one is:

8 x 8 x 7 x 7 x 7 x 7 = 1,322,496

Therefore, there are 1,322,496 possible 6-digit security numbers that can be made with the digits 1-9 without repeating any digits, where the first digit is not one.

c) To create an odd number, the last digit must be an odd number, which means we have 5 choices for the last digit (1, 3, 5, 7, or 9). For the first digit, we cannot use 0 or 1, so we have 7 choices. For the remaining 4 digits, we still have 8 choices for each digit (since we can use any digit).

Using the permutation formula again, the number of possible 6-digit security numbers with the last digit being odd is:

7 x 8 x 8 x 8 x 8 x 5 = 7,1680

Therefore, there are 7,1680 possible 6-digit security numbers that can be made with the digits 1-9 without repeating any digits, where the last digit is odd.

d) To create a number greater than 300,000, the first digit must be 3, 4, 5, 6, 7, 8, or 9. If the first digit is 3, we have 7 choices for the first digit (3, 4, 5, 6, 7, 8, or 9). For the remaining 5 digits, we still have 8 choices for each digit.

If the first digit is not 3, we have 6 choices for the first digit (since we cannot use 1 or 2). For the remaining 5 digits, we still have 8 choices for each digit.

Using the permutation formula again, the number of possible 6-digit security numbers greater than 300,000 is:

7 x 8 x 8 x 8 x 8 x 8 + 6 x 8 x 8 x 8 x 8 x 8 = 2,526,720

Therefore, there are 2,526,720 possible 6-digit security numbers that can be made with the digits 1-9 without repeating any digits, where the number is greater than 300,000.

e) To create a number greater than 750,000, the first digit must be 8 or 9. If the first digit is 8, we have 2 choices for the first digit (8 or 9). For the remaining 5 digits, we still have 8 choices for each digit.

If the first digit is 9, we only have one choice for the first digit (9). For the remaining 5 digits, we still have 8 choices for each digit.

Using the permutation formula again, the number of possible 6-digit security numbers greater than 750,000 is:

2 x 8 x 8 x 8 x 8 x 8 + 1 x 8 x 8 x 8 x 8 x 8 = 262,144

Therefore, there are 262,144 possible 6-digit security numbers that can be made with the digits 1-9 without repeating any digits, where the number is greater than 750,000.

answered
User Artefacto
by
7.9k points

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