Combinatorics Question?

Question

How many odd 4-digit numbers, all of the digits different, can be formed using the digits 0-9 if the digit 5 must be included?

Answer 1

There's 4 positions for the digit 5. Of each of the remaining 3 digits, only numbers 1,3,7,9 can be used, and but once. So, the total number possible is:

(4)(4)(3)(2) = 96 ways.

Addendum: Okay, let me revise that. Odd NUMBERS, not all odd digits. Let's see, if 5 was in the last digit, then there are

(8)(7)(6) = 336 ways to form a odd number with 5 at the last dight.

If 5 was in any of the other 3 positions, then we have the possible sequences of remaining even and odd digits (regardless of where the 5 is):

EEO = (4)(3)(4) = 48 ways
EOO = (4)(4)(3) = 48 ways
OEO = (4)(4)(3) = 48 ways
OOO = (4)(3)(2) = 24 ways

So, because the 5 can be one of 3 places, we have the total of

(3)(48+48+48+24) + 336 = 840 ways

Answer 2

For a number to be odd it must end in an odd number. Which means any of the first 4 digits could be even or odd. 5241 is odd, even thought it has even digits.

So that means we always have a 5, the number always ends in an odd digit and the other digits could be any number odd or even.

Let's look at all the possibilities we could have if 5 is the first digit.
There's only one way to have 5 as the first digit. That means we have 9 choices for the second, 8 for the third and finally all the remaining odd digits for the last. Only problem is, were the second and third digits odd or even?

You may want to think of the problem this way:
digit 1: 5 (1 choice)
digit 2: even (5choices)
digit 3: even (4 choices)
digit 4: odd (4 choices)
80 total choices
AND

digit1: 5 (1 choice)
digit 2: odd (4 choices)
digit 3: even (5 choices)
digit 4: odd (3 choices)
60 total choices

AND
digit1: 5 (1 choice)
digit 2: even (5 choices)
digit 3: odd (4 choices)
digit 4: odd (3 choices)
60 total choices

AND
digit 1: 5 (1 choice)
digit 2: odd (4 choices)
digit 3: odd (3 choices)
digit 4: odd (2 choices)
24 total choices

Add them all together: 224

Those are all the possible ways you can have an odd four digit number, with a 5 in it and no repetitions. Now all you have to do it look at what happens when you move the 5 to the second, third or fourth digit. You'll notice there are certain patterns, but combinatorics drives me slightly nutty, so I always have to organize it visually.

Answer 3

2 cases: 5 is the last digit, 5 isn't the last digit
Case 1: 8*8*7 = 448
Case 2: 4*C(8,2)*3!-14 = 658
There are 4 cases for the last digit. Besides the digit 5 and the last digit, we need 2 more digits which can be chosen C(8,2)=28 ways. There are 3!=6 ways to rearrange these 3 digits to form the first 3 digits. We must subtract the 14 cases where 0 is the first digit.

448+658=1106

Note that the digits do not have to be odd, just the last digit.

Answer 4

There are a total of 5 odd numbers from 0-9: One of them must be five, this leaves 4 that you can choose from for the reamaining 3. There are 4C3=4 combinations of choices that you can make; and for each combination of 4 numbers to work with, you have 4*3*2*1 ways in which you can arrange them. Therefore, there are a total of 4*24=96 different ways you can accomplish this task.