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
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
now, add them all up
(any questions?)