Use Pascal's Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1 (this is the row you use)
So, the seventh term of (3x + 2y)^9 will be 84xy
Let's solve your system by substitution.
y=2x+1;y=4x−1
Step: Solve y=2x+1for y:
y=2x+1
Step: Substitute2x+1foryiny=4x−1:
y=4x−1
2x+1=4x−1
2x+1+−4x=4x−1+−4x(Add -4x to both sides)
−2x+1=−1
−2x+1+−1=−1+−1(Add -1 to both sides)
−2x=−2
−2x
−2
=
−2
−2
(Divide both sides by -2)
x=1
Step: Substitute1forxiny=2x+1:
y=2x+1
y=(2)(1)+1
y=3(Simplify both sides of the equation)
Answer:
x=1 and y=3
When you have 3 choices for each of 6 spins, the number of possible "words" is
3^6 = 729
The number of permutations of 6 things that are 3 groups of 2 is
6!/(2!×2!×2!) = 720/8 = 90
A) The probability of a word containing two of each of the letters is 90/729 = 10/81
The number of permutations of 6 things from two groups of different sizes is
(2 and 4) : 6!/(2!×4!) = 15
(3 and 3) : 6!/(3!×3!) = 20
(4 and 2) : 15
(5 and 1) : 6
(6 and 0) : 1
B) The number of ways there can be at least 2 "a"s and no "b"s is
15 + 20 + 15 + 6 + 1 = 57
The probability of a word containing at least 2 "a"s and no "b"s is 57/729 = 19/243.
_____
These numbers were verified by listing all possibilities and actually counting the ones that met your requirements.
