Hello, any help will be helpful:
Thanks in advancea)Find the number of even 5-digit number using digits from 2-8 with no duplication.
b) How many of the above numbers contain a 4 and if the digit 2 is there, it must come last
i suppose 2 and 8 are included here.
to be even, the last digit must be 2,4,6 or 8. so there are 4 choices for the last digit. for the next digit, we have 6 choices (because there are 6 digits left), for the next, there are 5 choices (because we cannot choose something we already chose). continuing like this, we have the answer to be:
4(6)(5)(4)(3) = 1440
clarification here: what must come last?b) How many of the above numbers contain a 4 and if the digit 2 is there, it must come last
See if I count these right.
a): Since the number is even, it must end in 2,4,6,8
There are then 4 choices for the ending digit, but the other three can be distributed among the other spaces. Since there is not repetition, we have 6 choices for the fourth, 5 for the third, and so forth.
4(3*4*5*6)=1440
b): Place the 2 in thr last spot. Then the 4 can go in any of four spaces.
There are 60 choices for the other 3 spots. 4*60=240.