Question 1175384: Suppose a "combination" lock has the numbers 0 to 39 on it. In order to open the lock, a person must dial the correct three-number sequence, such as 12 left, 24 right, 6 left.
a. How many different three-number sequences for this lock are possible if all three numbers are even numbers? Assume repeated numbers are not allowed. (Note: zero is an even number.)
b. How many different three-number sequences are possible if only the middle number must be even? Assume repeated numbers are not allowed.
c. What would the answer to part (f) be if we assume repeated numbers are allowed?
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
(a) There are 20 odd and 20 even numbers on the lock. If all three numbers in the combination have to be even and all different, then there are...
20 choices for the first number;
then 19 choices for the second;
then 18 choices for the third.
By the fundamental counting principle, the total number of possible combinations is 20*19*18.
You can do the calculation....
(b) Work the problem in a similar manner....
However, the instructions are ambiguous. "...only the middle number must be even" might mean that the first and last numbers can be anything. But it might mean that ONLY the middle number is even, so that the first and last are odd.
(c) There is no part (f) in the question. But any similar problem would be worked in the same manner as (a) and (b).
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