# Permutations

• Nov 17th 2009, 06:22 PM
john-1
Permutations
How many four-digit numbers can you form with the digits 1, 2, 3, 4, 5, 6, and 7 if no digit is repeated?

How many of these four-digit numbers are odd numbers?

How many of them are even numbers?

I need help with the bolded questions. please explain, I have the answers from the textbook but I dont know how they got them.
• Nov 17th 2009, 06:41 PM
ANDS!
A. You have 7 items, and are meant to chose 4 of them to form a number. This is a permutation yes? How would you solve a permutation where you have N items and need to choose K?

B. This is a "tricky" one that requires a quick bit of reasoning. For a number among the amount you found in part A to be odd, it should have the LAST digit as the 1, 3, 5 or 7. Correct? Let one of those numbers be our base. We are then left with 6 numbers and we need to choose 3 to pair with our odd number. Do you see where I am going with this?

C. The same process as B.

Let me know what you get.
• Nov 17th 2009, 06:43 PM
Plato
Quote:

Originally Posted by john-1
1) How many four-digit numbers can you form with the digits 1, 2, 3, 4, 5, 6, and 7 if no digit is repeated?

2) How many of these four-digit numbers are odd numbers?

3) How many of them are even numbers?
.

#1 $(7)(6)(5)(4)$ WHY? You explain that!

#2 $(4)(6)(5)(4)$ WHY? You explain that!

#3 You do this one for yourself!
• Nov 17th 2009, 06:54 PM
john-1
b) 6P3 = 120 x 4 = 480

c) 6P3 =120 x 3 = 360

How's this look ANDS! ?

Plato:
how did you get that arrangement? 4 x 6 x 5 x 4?
• Nov 17th 2009, 06:56 PM
ANDS!
That's it. Just remember a lot of this stuff LOOKS more difficult than it really is. The toughest part of that problem is having that quick lightbulb go off and think "Oh heck, I know what an odd number looks like!"

Quote:

how did you get that arrangement? 4 x 6 x 5 x 4?
$(6)(5)(4)$ is the same is 6P3. The leading four is the fact that you have four odd-bases to choose from.