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.
Learn more about Sequences on:
brainly.com/question/12246947
#SPJ4
Answer:
a = 46
b = 20
c = 15
b+c ≤ a
20 + 15 ≤ 46
61 ≤ 46
Step-by-step explanation:
Answer:
Step-by-step explanation:
x^2+14x+45=0
x^2+(9+5)x+45=0
x^2+9x+5x+45
x(x+9)+5(x+9)=0
(x+9)(x+5)=0
either x+9=0 Or,x+5=0
x+9=0
x=0-9
x=-9
x+5=0
x=0-5
x=-5
therefore x=-9,-5
