If b is in the first position then c can be in any 1 of the remaining 6 positions.
If we start with ab then the letter c can be in any one of 5 positions and if we have aab there are 4 possible positions for c and so on.
So the total number of possible sequences where b comes first = 6+5+4+3+2+1 = 21.
The same argument applies when c comes before b so that gives us 21 ways also.
So the answer is 2 *21 = 42 different sequences.
A more direct way of doing this is to use factorials:-
answer = 7! / 5! = 7 * 6 = 42.
( We divide by 5! because of the 5 a's.)
The material for the lamp of the wiring would be cutting with pliers.
7 x 10^5
+ 2 x 10^4
+ 0 x 10^3
+ 0 x 10^2
+ 8 x 10^1
+ 0 x 10^0
Answer:
Infinitely Many Solutions.
Step-by-step explanation:
x+2y=0
6y=-3x
y=-3/6x
y=-1/2x
x+2(-1/2x)=0
x-2/2x=0
x-x=0
0=0
infinitely many solutions
