1. ## simple probability

how many even numbers can be formed with the digits 3,4,5,6,7 by using some or all of the numbers (repetitions are not allowed) ?

what i did was
3x4P4 + 3x4P3 + 3x4P2 + 3x4P1 + 3
but the answer was was wrong

2. Originally Posted by Jhevon
to be even, the number has to end with the 4 or the 6.

to end with 4, the number is of the form ****4, where the *'s represent the other numbers. there are 4! ways to make such a number, by permuting the 4 *'s

the same holds true if we end the number with 6.

so that in all, there are 2*4! = 48 even numbers
LOL thanks!. looks like i thought 3 was an even number hahahahaha.
problem solved

3. Originally Posted by z1llch
LOL thanks!. looks like i thought 3 was an even number hahahahaha.
problem solved
haha, i deleted my post. i misread the problem, i didn't see the "some" part. i assumed we needed to use all the numbers (too easy ).

stay tuned, i will update this post with the solution soon

EDIT: solution.

to be even, the number has to end with the 4 or the 6. but we can use all 5 of the numbers, or 4 or 3 or 2 or 1 of them.

okay. using all 5 numbers:

to end with 4, the number is of the form ****4, where the *'s represent the other numbers. there are 4! ways to make such a number, by permuting the 4 *'s

the same holds true if we end the number with 6.

so that in all, there are 2*4! = 48 even numbers

using 4 of the numbers:

***4 -----> 4*3*2 = 4! even numbers
***6 -----> 4*3*2 = 4! even numbers

so, 2*4! = 48 even numbers

using 3 of the numbers:

**4 ------> 4*3 = 12 even numbers
**6 ------> 4*3 = 12 even numbers

so, 2*12 = 24 even numbers

using 2 of the numbers:

*4 ------> 4 even numbers
*6 ------> 4 even numbers

so, 2*4 = 8 even numbers

using 1 of the numbers:

4 --------> 1 even number
6 --------> 1 even number

so, 2 even numbers