Question 1162297: The access code consists of four digits. The first digit cannot be zero and the last digit must be odd. How many different codes
are available? Digits cannot repeat.
Here's my work so far....
4 digits total, 1st cannot be 0, last digit must be odd, no repeats allowed
1st digit cannot be 0, with no repeats allowed leaves (9) options
2nd Digit no repeats allowed but could be 0 which leaves (9) options again
3rd Digit no repeats allowed, which leaves (8) options
4th Digit must be odd, no repeats, 1,3,5,7,9 this leaves (5) options
Question does not mention if previous digits were odd numbers which would change values quite a bit, but this teacher is straightforward, not into complicate multi answer questions leading me to stick with 5 options for 4th digit
This leads me to take values and multiply 9x9x8x5=3,240 possible combinations. I'm just not sure if this is right or if I'm missing something. Thanks for your help!
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
Last digit is one of 5 odd digits 1, 3, 5, 7 or 9. 5 options.
First digit is any of the remaining 9 digits except of 0. 9-1 = 8 options.
So far, 2 digits of the 10 digits are used; the remaining 8 digits of the 10 digits are still free.
Second digit from the left is any of remaining 8 digits. 8 options.
Third digit from the left is any of remaining 7 digits. 7 options.
In all, there are 5*8*8*7 = 2240 different combinations.
ANSWER. Under the given conditions, 2240 different codes are possible.
Solved.
I am ready to accept your "thanks" for my teaching.
|
|
|
