Question 1140802: Hi,
I have a question where I don't even know where to start. I'm not sure if it is a permutation or a combination, but I know what both of them mean. Here it is:
Find how many six-digit numbers can be formed from the digits 2, 3, 4, 5, 6 and 7 (with repetitions), if:
the numbers formed must be even, the numbers formed must be divisible by 25, and the odd digits must occupy even positions (i.e. 2nd, 4th and 6th) and the even digits must occupy odd positions (i.e. 1st, 3rd and 5th).
Thank you so much for your time, it really does mean a lot!
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
I read that as a single problem, with all those requirements at the same time.
In that case, the answer is simple: 0.
The requirements say that the number must be even and the last digit must be odd. There are no numbers that satisfy those conditions.
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