Re: Permutations question

Hey skg94.

For 1) can we have any number that is 1, 2, or 3 digits given the five possibilities? If this is the case, then I think you are underestimating the possibilities.

Re: Permutations question

I approached 1) by considering the quantity of single digit numbers that can be made using the five possibilities. That's just n_P_r where n=5 and r=1. I thought 5_P_r might represent the quantity of r-digit numbers that can be made using five possibilities (which it does) so I figured if I added the total quantity of numbers that can be made from r=1 to r=3, then I'd get the solution to your problem. That is:

5_P_1 + 5_P_2 + 5_P_3 which gave me 85.

I then noticed that for r=2: 01=1, 02=2, 03=3, 04=4 and for r=3: 012=12, 013=13, 014=14, 021=21, 023=23, 024=24, 031=31, 032=32, 034=34, 041=41, 042=42, 043=43. Since they represent the same number, these 16 possibilities must be discarded.

85-16=69

I'm not sure if this is the only way to arrive at the solution since I did out by hand all the possibilities which I thought would repeat (even though I knew each of n_P_r, n_P_(r+1) etc... for r>1 had distinct repeats). Perhaps if I figured out a more general rule that tells me how many repeats I have for each "r" when r>1, then I wouldn't have to list the possibilities for repeating numbers and in doing so, I would arrive with the complete solution.

Am I on the right track here?

Re: Permutations question

Hi Skg94

For (1), is it that you are not counting the number zero as a number, (your third component should be 5). ?

For (2), the possible forms for the four digits are,

(Odd, 5 or 9) (any one of 6) (any one of 5) (even, 0, 2, 6 or 8).................8 times 30 possibles.

(Even, 6 or 8) (any one of 6) (any one of 5) (even, 0, 2, one of (6 or 8))............6 times 30 possibles.

Of these, it looks like there are four basic forms ending with zero, 5**0, 9**0, 6**0 and 8**0, with 30 possibles for each.

Re: Permutations question

Hello, skg94!

Quote:

2. How many 4-digit even numbers greater than 5000

can you form using the digits 0,1,2,3,5,6,8 and 9 without repetitions?

There are 4 even digits: {0, 2, 6, 8} and 4 odd digits: {1, 3, 5, 9}.

The first digit can be {5, 6, 8, 9}.

If the first digit is odd: 5 or 9 . . . 2 choices

. . the last digit can be: 0, 2, 6, 8 . . . 4 choices

The other two digits have (6)(5) = 30 choices.

Hence, there are: (2)(4)(30) = **240** such numbers.

If the first digit is even: 2, 6, or 8 . . . 3 choices

. . the last digit has 3 choices . . . 3 choices

The other two digits have (6)(5) = 30 choices.

Hence, there are: (3)(3)(30) = **270** such numbers.

Therefore, there are: $\displaystyle 240 + 270 \,=\,\boxed{510}$ even numbers greater than 5000.

Quote:

b) How many of these numbers end in 0?

The last digit is 0 (zero) . . . 1 choice.

The first digit is 5, 6, 8, or 9 . . . 4 choices

The other two digits have (6)(5) = 30 choices.

Therefore, there are: $\displaystyle (1)(4)(30) = \boxed{120}$ such numbers.

1 Attachment(s)

Re: Permutations question

Hi Skg94,

Counting problems are hard -- don't let anyone tell you otherwise. As illustration, the answer to 2 given by Soroban is wrong (he did come up with the right number for the second part).

Any 4 digit number is of the form 1000d + 100c + 10b + a. Choose the digits in the order d, a, b, c.

For d=5, there are 4*6*5 = 120 choices

For d=7, there are 4*6*5 = 120 choices

For d=6, there are 3*6*5 = 90 choices (when d=6, the only choices for a are 0, 4 and 8)

For d=8, there are 90 choices.

So a grand total of 240+180 = 420.

Exactly similarly, you can count the number with 0 as the least significant digit and get 120.

If you're a computer programming whiz, you can check answers to such "small" problems with a little computer program. I did so for this problem and generated all the 4 digit integers in the question. If you know programming, I'd be happy to send you the program as a private message. See the attachment.

Attachment 26315