How many 7 digit numbers can be formed using the numbers 3, 4, 5, 6, 7, 8, and 9 if the last digit must be prime and repetition is not allowed?

Question

How many 7 digit numbers can be formed using the numbers 3, 4, 5, 6, 7, 8, and 9 if the last digit must be prime and repetition is not allowed?

 

How does one go about answering this? Thanks.

Answer 1

First, let's figure out how many prime numbers there are from 3-9. They are 3, 5, and 7. So there are 3 possibilities for the last digit.

 

For the first 6 digits, we will be choosing the permutation of the 6 remaining digits. It's a permutation and not a combination because order matters (123 and 321 are not the same thing).  

 

6P6

= 6!/(6-6)!

= 6!/1

= 720

 

So 720 possibilities for the first 6 digits and 3 possibilities for the last digit.  That means the total number of possible numbers would be 720 * 3 = 2160.

 

Note that if repetition of numbers had been allowed, then the answer would have changed dramatically. In that case, the first 6 digits would each have 7 unique possibilities. So the total would increase to 7^6 * 3 = 352,947 possible numbers.