Question 1182760: In how many ways can the numbers 1,2,3,4,5,6 be arranged in a row so that the product of any 2 adjacent numbers is even?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
These arrangements are those and only those, where for each odd number its closest neighbors are even numbers.
In other words, these arrangements are those, where odd and even numbers occupy alternate positions.
There are 3! = 6 arrangements for the three ODD numbers in such permutations, and 3! = 6 arrangements
for the three EVEN numbers in such permutations.
Do not forget to include factor 2 to account to which number, - an odd or an even, - goes first.
So, the ANSWER is: there are 2*6*6 = 72 such arrangements.
Solved.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The arrangements must not have two odd numbers next to each other. With three odd numbers and three even numbers, that means the six numbers must alternate between even and odd.
So in terms of even and odd numbers, the arrangement must be of the form
ABABAB
The first number can be any of the 6.
The second number must be one of the 3 of the opposite parity.
The third must be one of the remaining 2 that have different parity than the second.
The fourth must be one of the remaining 2 that have different parity than the third.
The fifth must be the 1 that has different parity than the fourth.
There is only 1 left for the sixth position.
Total number of arrangements with alternating even and odd: 6*3*2*2*1*1 = 72.
ANSWER: 72
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