Amy's garage door code consists of 4 digits. The code allows any of the digits 0 through 9 inclusive, but the last digit must not be a 1.
The difference between the number of codes that are possible if repeats are allowed and the number of codes that are possible if repeats are not allowed is ___
5 years ago
Answered By Leonardo F
We can start with the possibility of repetition between the numbers. If the code has 4 digits, and the last one cannot be 1, between 0 and 9 we have 10 numbers. If repetitions are allowed, the possibilities for the digits are:
10 x 10 x 10 x 9 = 9000 codes if repetitions are allowed
The number 9 is there because we cannot use the number 1 for the last digit.
For the case in which repetitions are now allowed, we have to start with the restriction, the last digit. It cannot be number 1, so we have 9 possibilities for it. For the first digit, we are left with 9 possibilities, because we can use the number 1, but we cannot use the number that we chose for the last digit. For the other two digits, we lose one possibility for each one, because we cannot repeat the numbers. Hence, the calculation is as follows:
9 x 8 x 7 x 9 = 4356 codes if repetitions are not allowed
Hence, the difference will be:
9000 - 4356 = 4644
Final answer: 4644
5 years ago
Answered By Leonardo F
Actually, I made a mistake. The first number is 4536, not 4356. Hence, the final answer will be 4464.
5 years ago
Answered By Mohammed B
First draw 4 lines to represent the 4 digits.
Each line represents a digit of the 4 digit passcode.
The first line has 10 possible numbers (count from 0-9)
With repeats: You have 10×10×10x9 possible pass codes or 9000 passcodes. (The first-third line has 10 possible numbers since it can repeat and be from 0-9,whereas the last line has 9 possible numbers since it CANNOT be 1)
Without repeats: You have 10×9×8×6=4320 possible pass codes. (The first line has 10 possible numbers (0-9), second line has 9 possible numbers (this is not 10 like the previous question since the numbers cannot repeat and 1 must be deducted from 10), third line has 8 possible numbers, and the fourth line has 6 possible numbers (since the numbers cannot repeat and the last digit cannot equal 1 so 2 is deducted from 8)
Difference:9000-4320=4680
5 years ago
Answered By Obinna E
Case 1: Repition of numbers allowed
If the code has 4 digits, and the last one cannot be 1, between 0 and 9 we have 10 numbers. If repetitions are allowed, the possibilities for the digits are:
10 x 10 x 10 x 9 = 9000 codes
The number 9 is there because we cannot use the number 1 for the last digit.
Case 2: Repition of numbers not allowed & 1 must not be the last digit
For the case in which repetitions are now allowed, we have:
10 x 9 x 8 x 7 = 5040 codes if repetitions are not allowed
If last digit must not be 1, we have 3 digit spaces & 9 number options left. This means that the case for repition of numbers not allowed & 1 must not be the last digit is:
5040 - (3 * 9) = 5013 options
Hence, the difference will be:
9000 - 5013 = 4644
Final answer: 3987
5 years ago
Answered By Majid B
1) Repetition is allowed
Consider four empty boxes as four digits. Each empty box can be filled with 10 options ("zero" to "ten") except the last box that has only 9 options ("one" is not allowed for the last digit).
Then according to Counting Principle, the all possible options are:
10 * 10 * 10 * 9 = 9000
2) Repetition is not allowed:
First, calculate the number of options that "one" is at the last box. In this case, we have only 3 empty boxes that have to be filled with 9 numbers (because we already used "one"). Because repetition is not allowed, the number of options for filling these 3 boxes are:
9 * 8 * 7 = 504
Now, consider all options (including "one" at the last box) when repetition is not allowed:
10 * 9 * 8 * 7 = 5040
Therefore, the number of options when repetition is not allowed and the last digit cannot be "one" is calculated as:
5040 - 504 = 4536
3) Difference
Finally, the final answer is:
9000 - 4536 = 4464
5 years ago
Answered By Sara-Jeanne V
1) Repetition is allowed
Facts: Four digit code, last cannot be 1
Look at a 2 digit code where 1st can be 3 or 5 and second can be 6, 5 or 8. Then the possibles are 36, 35, 38, 56, 55, and 58, right? Then that is 6 possible combinations. Notice that that is 2x3=6. There were two possibilities for the first digit and 3 for the second.
Therefor your problem is 4 codes; 3 can have 0-9 and the last can have 0, or 2-9. Then the formula is 10x10x10x9 which is 9000.
2) Repeats not allowed
Facts: Four digit code, last cannot be 1, no two digits in a row.
Let's start by ingoring the "last cannot be 1" rule. Then we have the first digit (call it "a") can be 0-9, the sencond digit ("b") can be 0-9 but not a, the third digit ("c") can be 0-9 but not a or b, and the fourth digit ("d") can be 0-9 but not a, b or c. So 10x9x8x7 is 5040.
Now we need to remove the combinations where the last digit is 1. These combinations have
the fourth digit ("d") can only be 1,
the third digit ("c") can be 0-9 but not be 1,
the sencond digit ("b") can be 0-9 but not c or 1,
the first digit (call it "a") can be 0-9 but not b, c or 1
So 1*9*8*7 is 504.
So the total of norepeating combinations that do not end in 1 are 5040-504 is 4536.
Final Answer: The number of combinations with repititions is 9000 and w/o repetitions is 4536 so the difference is 4464.
5 years ago
Answered By Leonardo F
We can start with the possibility of repetition between the numbers. If the code has 4 digits, and the last one cannot be 1, between 0 and 9 we have 10 numbers. If repetitions are allowed, the possibilities for the digits are:
10 x 10 x 10 x 9 = 9000 codes if repetitions are allowed
The number 9 is there because we cannot use the number 1 for the last digit.
For the case in which repetitions are now allowed, we have to start with the restriction, the last digit. It cannot be number 1, so we have 9 possibilities for it. For the first digit, we are left with 9 possibilities, because we can use the number 1, but we cannot use the number that we chose for the last digit. For the other two digits, we lose one possibility for each one, because we cannot repeat the numbers. Hence, the calculation is as follows:
9 x 8 x 7 x 9 = 4356 codes if repetitions are not allowed
Hence, the difference will be:
9000 - 4356 = 4644
Final answer: 4644
5 years ago
Answered By Leonardo F
Actually, I made a mistake. The first number is 4536, not 4356. Hence, the final answer will be 4464.
5 years ago
Answered By Mohammed B
First draw 4 lines to represent the 4 digits.
Each line represents a digit of the 4 digit passcode.
The first line has 10 possible numbers (count from 0-9)
With repeats: You have 10×10×10x9 possible pass codes or 9000 passcodes. (The first-third line has 10 possible numbers since it can repeat and be from 0-9,whereas the last line has 9 possible numbers since it CANNOT be 1)
Without repeats: You have 10×9×8×6=4320 possible pass codes. (The first line has 10 possible numbers (0-9), second line has 9 possible numbers (this is not 10 like the previous question since the numbers cannot repeat and 1 must be deducted from 10), third line has 8 possible numbers, and the fourth line has 6 possible numbers (since the numbers cannot repeat and the last digit cannot equal 1 so 2 is deducted from 8)
Difference:9000-4320=4680
5 years ago
Answered By Obinna E
Case 1: Repition of numbers allowed
If the code has 4 digits, and the last one cannot be 1, between 0 and 9 we have 10 numbers. If repetitions are allowed, the possibilities for the digits are:
10 x 10 x 10 x 9 = 9000 codes
The number 9 is there because we cannot use the number 1 for the last digit.
Case 2: Repition of numbers not allowed & 1 must not be the last digit
For the case in which repetitions are now allowed, we have:
10 x 9 x 8 x 7 = 5040 codes if repetitions are not allowed
If last digit must not be 1, we have 3 digit spaces & 9 number options left. This means that the case for repition of numbers not allowed & 1 must not be the last digit is:
5040 - (3 * 9) = 5013 options
Hence, the difference will be:
9000 - 5013 = 4644
Final answer: 3987
5 years ago
Answered By Majid B
1) Repetition is allowed
Consider four empty boxes as four digits. Each empty box can be filled with 10 options ("zero" to "ten") except the last box that has only 9 options ("one" is not allowed for the last digit).
Then according to Counting Principle, the all possible options are:
10 * 10 * 10 * 9 = 9000
2) Repetition is not allowed:
First, calculate the number of options that "one" is at the last box. In this case, we have only 3 empty boxes that have to be filled with 9 numbers (because we already used "one"). Because repetition is not allowed, the number of options for filling these 3 boxes are:
9 * 8 * 7 = 504
Now, consider all options (including "one" at the last box) when repetition is not allowed:
10 * 9 * 8 * 7 = 5040
Therefore, the number of options when repetition is not allowed and the last digit cannot be "one" is calculated as:
5040 - 504 = 4536
3) Difference
Finally, the final answer is:
9000 - 4536 = 4464
5 years ago
Answered By Sara-Jeanne V
1) Repetition is allowed
Facts: Four digit code, last cannot be 1
Look at a 2 digit code where 1st can be 3 or 5 and second can be 6, 5 or 8. Then the possibles are 36, 35, 38, 56, 55, and 58, right? Then that is 6 possible combinations. Notice that that is 2x3=6. There were two possibilities for the first digit and 3 for the second.
Therefor your problem is 4 codes; 3 can have 0-9 and the last can have 0, or 2-9. Then the formula is 10x10x10x9 which is 9000.
2) Repeats not allowed
Facts: Four digit code, last cannot be 1, no two digits in a row.
Let's start by ingoring the "last cannot be 1" rule. Then we have the first digit (call it "a") can be 0-9, the sencond digit ("b") can be 0-9 but not a, the third digit ("c") can be 0-9 but not a or b, and the fourth digit ("d") can be 0-9 but not a, b or c. So 10x9x8x7 is 5040.
Now we need to remove the combinations where the last digit is 1. These combinations have
the fourth digit ("d") can only be 1,
the third digit ("c") can be 0-9 but not be 1,
the sencond digit ("b") can be 0-9 but not c or 1,
the first digit (call it "a") can be 0-9 but not b, c or 1
So 1*9*8*7 is 504.
So the total of norepeating combinations that do not end in 1 are 5040-504 is 4536.
Final Answer: The number of combinations with repititions is 9000 and w/o repetitions is 4536 so the difference is 4464.