2 Questions about Counting & Permutations

  • #1
139
0

Homework Statement



How many eight-digit numbers can be formed under each condition?

a) The leading digit cannot be zero, the fifth digit cannot be 6 or 8, and the number must be less then 75,000,000.

b) The leading digit cannot be zero, the number must be divisible by 5, the fourth digit cannot be 2, and no repetition of digits is allowed.


Homework Equations



-None-

The Attempt at a Solution



So because you have 8 spaces, and any number between 0-9 can fit, you have a total of 10 choices you can put in a single spot.

Part A:
_ _ __ __ _ __ __ __
7 5 10 10 8 10 10 10

Multiplied together = 28,000,000 different eight-digit combinations.

The number has to be lower then 75,000,000 so highest number you can have there is a 7, meaning 8, but no 0 either, so 7. The next number can only be up to a 4, so you have 5 choices between 0-4. The next two spots are 10's because they have no parameters. The fifth digit is a 8 because you can't have the numbers 6 or 8 there. And the rest are 10's.

I tried doing this based of the number: 74,999,999 so it would be less then 75,000,000 and fill the rest based off of that.

Part B:

_ _ _ _ _ _ _ _
1 9 8 6 6 5 4 3

Multiplied together = 155,520 different choices

For the first digit, it can't be a 0, so it takes the choices for that spot to 1-9, and the number has to be divisible by 5, so there is basically only one choice, of 5, then the next number is 9 because no repetition is allowed and same for the 8. The fourth digit can't be a 2, so instead of 7, it lowers to 6. The next digit is 6 too because you still have that many choices. And then it decreases one each time because of the no repetition.

I think I did them right, but I am not sure, that is why I am looking for help. Thanks!
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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A: If the first digit is 7, then the second digit must be one of 0, 1, 2, 3, or 4. But if the first digit is less than 7, the second digit may be 0 to 9.

So you need to do this is two parts. How many such numbers are the with first digit 7? How many such numbers are the with the first digit less than 7? Add those together.

B: If the number is divisible by 5 there are two possible last digits- 5 or 0, not just 5.
 
  • #3
139
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Oh so for a) you would have to have two circumstances and add them?

Like the one I got: 28,000,000 plus the new one?

If so, that means the other circumstance would have to be:

_ __ __ __ _ __ __ __
7 10 10 10 8 10 10 10

Multiplied together = 56,000,000

Answer to A) Total of 84,000,000 different combinations

For b) it was talking about the whole number? I thought it was talking about the first number. I know for a number to be divisible by 5, it ends in 0 or 5.

So the new answer would be:

_ _ _ _ _ _ _ _
9 9 8 6 6 5 4 3

Answer for B) 1,399,680 different combinations

Is this correct now?
 
  • #4
139
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Bumping. If anyone can help.
 
  • #5
157
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a) I believe you would calculate the number of combinations with 7 as the leading number first, then add the number of combinations with the leading number as 6 or lower.

b) You have the last number wrong. There are only 2 possibilities for the last number, not 3. Change that and you should get the correct number of combinations.
 
  • #6
139
0
Ok, thanks, I tried that and I finally got it. Do you have aim? I have a few questions about some other problems, and I think it is a faster way to communicate. If so, can you please inbox it to me?
 
  • #7
157
0
Sorry, I don't have AIM.
 

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