65 sequences.
Lets solve the problem,
The last term is 0.
To form the first 18 terms, we must combine the following two sequences:
0-1 and 0-1-1
Any combination of these two sequences will yield a valid case in which no two 0's and no three 1's are adjacent
So we will combine identical 2-term sequences with identical 3-term sequences to yield a total of 18 terms, we get:
2x + 3y = 18
Case 1: x=9 and y=0
Number of ways to arrange 9 identical 2-term sequences = 1
Case 2: x=6 and y=2
Number of ways to arrange 6 identical 2-term sequences and 2 identical 3-term sequences =8!6!2!=28=8!6!2!=28
Case 3: x=3 and y=4
Number of ways to arrange 3 identical 2-term sequences and 4 identical 3-term sequences =7!3!4!=35=7!3!4!=35
Case 4: x=0 and y=6
Number of ways to arrange 6 identical 3-term sequences = 1
Total ways = Case 1 + Case 2 + Case 3 + Case 4 = 1 + 28 + 35 + 1 = 65
Hence the number of sequences are 65.
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Answer:
-3e/2
Step-by-step explanation:
Its D because when you use some graph paper point A and point B are 6 units away from each other
Answer:
28
Step-by-step explanation:
Here is the complete question used in answering this question :
Three-fourths of the boats in the marina are white. 4/7 of the remaining boats are blue , and the rest are red . If there are 9red boats, how many boats are in the marina?
We first have to determine the fraction of the boats that are red
1 - (
+ 
)
1 - 
1 - 
1 - 1
= 9/28
so, 9/28 of the boats are red
let b represent the total number of boats
9/28 x b = 9
to find b, divide both sides of the equation by 28/9
b = 28
Answer:
f=7x g=3 x=18
Step-by-step explanation:
